C++博客-Brian Warehouse-随笔分类-POJ

POJ 1046 Color Me Less

Brian — Tue, 17 Aug 2010 17:00:00 GMT

唉，半年没有切题的后果就是这道水题写了整整3个小时。。。

题目的要求是输入若干组颜色数据，前16组是目标组，后面剩下的都是用来尝试与其映射的，用后面与前面的一次匹配，各循环16次，各自最小的D的相应映射组就是我们要的结果。

C++ 编译器，220K 0MS

#include<iostream>
#include<climits> // 为第一组数据运算做的约束
using namespace std;
int RGB[16][3],in[3],out[3]; // in 用来接收数据, out 用来暂存映射正确的数据

int main(){
    for (int i=0; i<16; i++) // 输入的数据中前16组是 target set
        cin>>RGB[i][0]>>RGB[i][1]>>RGB[i][2];
    while (1) {
        cin>>in[0]>>in[1]>>in[2]; // 开始接收映射组
        if (in[0] == -1) break;
        int MIN=INT_MAX; // 2147483647
        for (int i=0; i<16; i++) {
            int sum=(RGB[i][0]-in[0])*(RGB[i][0]-in[0])+
                (RGB[i][1]-in[1])*(RGB[i][1]-in[1])+
                (RGB[i][2]-in[2])*(RGB[i][2]-in[2]); // 不需要开方，开方易产生误差
            if (sum < MIN) {
                out[0]=RGB[i][0];
                out[1]=RGB[i][1];
                out[2]=RGB[i][2];
                MIN=sum; // 最小的即映射成功
            }
        }
        cout<<"("<<in[0]<<","<<in[1]<<","<<in[2]<<") maps to ("
            <<out[0]<<","<<out[1]<<","<<out[2]<<")"<<endl;
    }
    return 0;
}

有一点要说明的是头文件，建议MSDN一下，你会发现很多有用的常量，拿来就能用。

Brian 2010-08-18 01:00 发表评论

POJ 1045 Bode Plot

Brian — Tue, 17 Aug 2010 06:49:00 GMT

我觉得有必要把它翻译一下，然后就会发现考得其实是数学，你在google 翻译上是得不到如下翻译的：
Description
考虑下面的交流电路。我们将假定电路在稳态。因此，节点1的电压和节点2的电压分别是v1 = VS coswt 和 v2 = VRcos (wt + q),其中Vs是电源电压，w是频率(弧度每秒)，t是时间。VR是电阻R两端电压下降的幅度，q是它的相位。

你需要写一个程序，以确定不同的w对应的VR值。您将需要两个电学定律来解决这个问题。第一
个是是欧姆定律，表述为V2 = iR，其中i是在电路顺时针流向的电流大小。
第二个是i = C d/dt (v1-v2)，i与电容器两板上的电压有关。"d/dt" 意为求关于t的求导。
Input
输入包括一行或多行。第一行包括三个实数和一个非负整数。实数按顺序是VS，R，C。
整数n是测试用例的个数。接下来的n行就是输入，要求一行一个实数，代表w的值。

Output
输出n行的VR值，注意，结果精确到小数点后三位。
下面需要推导一下求VR的公式：
V2=iR=CR d/dt (VS*cos(wt)-VR*cos(wt+q))=VRcos(wt+q)
= CR w (sin(wt+q)-sin(wt))=VRcos(wt+q)
下面用到高中数学当中的计算方法，分别令 t=0 和 wt+q=0 ，得到 CRw tan b = 1 和 VR=CRw VS sin b ，
然后利用三角函数中的万能公式，求得：VR = CRw VS / sqrt (1+ (CRw) ^ 2 ))

// C 编译器:
#include <stdio.h>
#include <math.h>
int main()
{
    int i=0,n;
    double VR,VS,R,C,w;
    scanf("%lf%lf%lf%d",&VS,&R,&C,&n);

    for (; i<n; i++)
    {
        scanf("%lf",&w);
        VR=C*R*w*VS / sqrt(1+C*C*R*R*w*w);
        printf("%.3lf\n",VR);
    }
    return 0;
}

注意 , 用 double

Brian 2010-08-17 14:49 发表评论

大数问题学习周(2)：POJ Grids 2980 大整数乘法

Brian — Tue, 17 Aug 2010 06:24:00 GMT

#include <iostream>
#include <string>
using namespace std;

int main()
{
    int a[200],b[200],c[400]={0},i,j,ls1,ls2;
    string s;

    for (cin>>s,ls1=s.length(),i=ls1-1,j=0; i>=0; i--)
        a[j++]=s[i]-'0'; //将第一个数逆序放入a数组

    for (cin>>s,ls2=s.length(),i=ls2-1,j=0; i>=0; i--)
        b[j++]=s[i]-'0'; //将第二个数逆序放入b数组

    for (i=0; i<ls1; i++)
        for (j=0; j<ls2; j++)
        {
            c[i+j] += a[i]*b[j];
            if(c[i+j] >= 10)
            {
                c[i+j+1] += c[i+j]/10;
                c[i+j] %= 10;
            }
        }

    i=399;
    while (i--)
        if (c[i]) break; //跳过所有前导0
    for (; i>=0; i--)
        printf("%d",c[i]); //输出主体部分
    return 0;
}

Brian 2010-08-17 14:24 发表评论

大数问题学习周(1)：POJ Grids 2981 大整数加法

Brian — Tue, 17 Aug 2010 06:21:00 GMT

#include <iostream>
#include <string>
using namespace std;
int main()
{
    int a[201]={0},b[200]={0},i,j,len,ls1,ls2,f=0; // 相加后结果放在a内
    string s;

    for (cin>>s,ls1=s.length(),i=ls1-1,j=0; i>=0; i--)
        a[j++]=s[i]-'0'; //将第一个数逆序放入a数组

    for (cin>>s,ls2=s.length(),i=ls2-1,j=0; i>=0; i--)
        b[j++]=s[i]-'0'; //将第二个数逆序放入b数组

    for (i=0,len=ls1>ls2?ls1:ls2; i<len; i++) // 注意len取二者较大的值
    {
        a[i] += b[i]; //相加结果放入a数组
        if (a[i] >= 10)
        {
            a[i] %= 10; //进位处理
            a[i+1]++;
        }
    }

    if (a[len]) printf("%d",a[len]); //所谓的前导0
    for (i=len-1; i>=0; i--)
        printf("%d",a[i]); //输出主体部分
    return 0;
}

Brian 2010-08-17 14:21 发表评论

POJ-grids 2808 校门外的树

Brian — Tue, 17 Aug 2010 06:19:00 GMT

题目描述

某校大门外长度为L的马路上有一排树，每两棵相邻的树之间的间隔都是1米。我们可以把马路看成一个数轴，马路的一端在数轴0的位置，另一端在L的位置；数轴上的每个整数点，即0，1，2，……，L，都种有一棵树。
马路上有一些区域要用来建地铁，这些区域用它们在数轴上的起始点和终止点表示。已知任一区域的起始点和终止点的坐标都是整数，区域之间可能有重合的部分。现在要把这些区域中的树（包括区域端点处的两棵树）移走。你的任务是计算将这些树都移走后，马路上还有多少棵树。

输入

输入的第一行有两个整数L（1 <= L <= 10000）和 M（1 <= M <= 100），L代表马路的长度，M代表区域的数目，L和M之间用一个空格隔开。接下来的M行每行包含两个不同的整数，用一个空格隔开，表示一个区域的起始点和终止点的坐标。

输出

输出包括一行，这一行只包含一个整数，表示马路上剩余的树的数目。

样例输入

样例输出

基本思路：将所有的树做标记，移走则标记为0，存在标记为1.由于不好判断给定数目是多少，故用向量。效率虽然不高，但是可以AC。

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int L,M,i=0,j,start,end,count=0;
    scanf("%d%d",&L,&M);
    vector<int> Mark(L+1,1);
    
    for (; i<M; i++) 
    {
        scanf("%d%d",&start,&end);
        for (j=start; j<=end; j++)
            Mark[j]=0;
    }
    for (i=0; i<=L; i++)
        if (Mark[i]==1)
            count++;
    printf("%d\n",count);
    return 0;
}

Brian 2010-08-17 14:19 发表评论

POJ Grids 2798 2进制转化为16进制

Brian — Tue, 17 Aug 2010 06:18:00 GMT

Description

输入一个2进制的数，要求输出该2进制数的16进制表示。
在16进制的表示中，A-F表示10-15

Input

第1行是测试数据的组数n，后面跟着n行输入。每组测试数据占1行，包括一个以0和1组成的字符串，字符串长度至少是1，至多是10000

Output

n行，每行输出对应一个输入。

Sample Input

2
100000
111

Sample Output

20
7

09小孩问我的一道题，原来写的代码足足有90多行，今天重写：

 
#include <iostream>
#include <string>
using namespace std;

int main()
{
    int n,pos,sec,i,j,w[4] = {1,2,4,8}; //sec是分段处理,pos是对应权值位置
    char x[17] = "0123456789ABCDEF"; //打表        
    string bin; //输入的二进制字符串
    cin>>n;
    while (n--)
    {
        cin>>bin;
        sec=bin.length()%4;
        pos=0;
        for (i=sec; i>0; i--)
            if (bin[sec-i]=='1')
                pos += w[i-1];
        if (sec) printf("%c",x[pos]);
        for (i=sec; i<bin.length(); i+=4)
        {
            pos=0;
            for (j=0; j<4; j++)
                if (bin[i+j]=='1')
                    pos += w[3-j];
            printf("%c",x[pos]);
        }
        printf("\n");
    }
    return 0;
}

Brian 2010-08-17 14:18 发表评论

POJ Grids 2745 显示器

Brian — Tue, 17 Aug 2010 06:17:00 GMT

Description

你的一个朋友买了一台电脑。他以前只用过计算器，因为电脑的显示器上显示的数字的样子和计算器是不一样，所以当他使用电脑的时候会比较郁闷。为了帮助他，你决定写一个程序把在电脑上的数字显示得像计算器上一样。

Input

输入包括若干行，每行表示一个要显示的数。每行有两个整数s和n (1 <= s <= 10, 0 <= n <= 99999999)，这里n是要显示的数，s是要显示的数的尺寸。

如果某行输入包括两个0，表示输入结束。这行不需要处理。

Output

显示的方式是：用s个'-'表示一个水平线段，用s个'|'表示一个垂直线段。这种情况下，每一个数字需要占用s+2列和2s+3行。另外，在两个数字之间要输出一个空白的列。在输出完每一个数之后，输出一个空白的行。注意：输出中空白的地方都要用空格来填充。

Sample Input

2 12345
3 67890
0 0

Sample Output

      --   --        --
|    |    | |  | |
|    |    | |  | |
--   --   --   --
| |       |    |    |
| |       |    |    |
--   --        --
---   ---   ---   ---   ---
|         | |   | |   | |   |
|         | |   | |   | |   |
|         | |   | |   | |   |
---         ---   ---
|   |     | |   |     | |   |
|   |     | |   |     | |   |
|   |     | |   |     | |   |
---         ---   ---   ---

Hint

数字(digit)指的是0，或者1，或者2……或者9。
数(number)由一个或者多个数字组成。

这题横跨我一个学期之久，后来无奈还是百度了一下：

#include <iostream>
using namespace std;

char n1[] = "- -- -----";
char n2[] = "|   ||| ||";
char n3[] = "|||||  |||";
char n4[] = "  ----- --";
char n5[] = "| |   | | ";
char n6[] = "|| |||||||";
char n7[] = "- -- -- --";

int main()
{
    int m;
    char n[10];
    cin>>m>>n;
    while ((n[0]-'0')||m)
    {
        int len = strlen(n);
        for (int i=0;i<len;i++)
        {
            cout<<" ";
            int num = n[i]-'0';
            for (int j=0;j<m;j++)
            {
                cout<<n1[num];
            }
            cout<<"  ";
        }
        cout<<endl;
        int temp = m;
        while (temp--)
        {
            for (int i=0;i<len;i++)
            {
                int num = n[i]-'0';
                cout<<n2[num];
                for (int j=0;j<m;j++)
                {
                    cout<<" ";
                }
                cout<<n3[num];
                cout<<" ";
            }
            cout<<endl;
        }
        for (int i=0;i<len;i++)
        {
            cout<<" ";
            int num = n[i]-'0';
            for (int j=0;j<m;j++)
            {
                cout<<n4[num];
            }
            cout<<"  ";
        }
        cout<<endl;
        temp = m;
        while (temp--)
        {
            for (int i=0;i<len;i++)
            {
                int num = n[i]-'0';
                cout<<n5[num];
                for (int j=0;j<m;j++)
                {
                    cout<<" ";
                }
                cout<<n6[num];
                cout<<" ";
            }
            cout<<endl;
        }
        for (int i=0;i<len;i++)
        {
            cout<<" ";
            int num = n[i]-'0';
            for (int j=0;j<m;j++)
            {
                cout<<n7[num];
            }
            cout<<"  ";
        }
        cout<<endl<<endl;

        cin>>m>>n;
    }
    return 0;
}

我想，第一次做出来这题的人真是了不起！

Brian 2010-08-17 14:17 发表评论

POJ 1008 Maya Calendar

Brian — Tue, 17 Aug 2010 06:16:00 GMT

思路：先求第一个的天数，然后用这个天数求第二个的表示方式，个人觉得不是水题

#include<stdio.h>
#include<string.h>
char hm[19][7]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax",
                "zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
char tm[20][9]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok",
                "chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};
int main()
{
    int i,m,n,day,year;
    char month[9];
    scanf("%d",&n);
    printf("%d\n",n);

    while (n--)
    {
        scanf("%d.%s%d",&day,month,&year);
        for(i=0; i<19; i++)
            if(strcmp(hm[i],month) == 0)
            {
                m=365*year+20*i+day;
                break;
            }
        printf("%d %s %d\n",m%260%13+1,tm[m%20],m/260);
    }
    return 0;
}

Brian 2010-08-17 14:16 发表评论

POJ 1007 DNA sorting

Brian — Tue, 17 Aug 2010 06:12:00 GMT

此程序耗费我尽3个小时之久，原因是做题前的规划没做好，一直没有想到整体排序的好办法，最后还是用了注意匹配的方法才解决了问题，我不知道为什么用冒泡不行，第一个字符串总是乱码。我觉得整体思路还是比较清晰的，只是方法可能有点傻，效率还行。
C 编译器 : 172K 0MS

#include <stdio.h>
#include <string.h>

typedef struct DNA
{
    char str[50]; // 存储字符串
    int count[2]; // [0] [1]都存放串的逆序数
}DNA;              // [1]中作为参考,用来和排序后的[0]匹配

int main()
{
    int i=0,j,k=0,n,m,temp;
    DNA or[100];
    scanf("%d%d",&n,&m);

    while (k<m) //获得数据并求各自逆序数
    {
        scanf("%s",&or[k].str);
        or[k].count[0]=0; // 此步不能忘
        for (i=0; i<n; i++)
            for (j=i+1; j<n; j++)
                if (or[k].str[i] > or[k].str[j])
                    or[k].count[0]++;
        k++;
    }

    for (i=0; i<m; i++)
        or[i].count[1]=or[i].count[0]; // 原逆序数存放顺序

    for (i=1; i<m; i++) // 对于各组串的逆序数进行排序,count[0]内容已打乱
    {
        k=i-1;
        for (j=i; j<m; j++)
            if (or[j].count[0] < or[k].count[0])
                k=j;

        temp=or[i-1].count[0];
        or[i-1].count[0]=or[k].count[0];
        or[k].count[0]=temp;
    }                // 这是典型的选择排序,只是对[0]单元的处理,稳定与否没关系


    for (i=0; i<m; i++)
        for (j=0; j<m; j++)
            if (or[i].count[0] == or[j].count[1]) // [0] 和 [1] 中逐一相比较
            {
                or[j].count[1]=-1; // 此步是相等时顺序不变的保证,相当于做了访问标记!
                printf("%s\n",or[j].str);
            }

    return 0;
}

Brian 2010-08-17 14:12 发表评论

POJ 1006 Biorhythms

Brian — Tue, 17 Aug 2010 06:10:00 GMT

典型的阅读理解题 , 读懂意思基本上思路就出来了，恰巧又是一道中文题，这里用枚举，其他不解释。


#include <stdio.h>
int main()
{
    int i,a,b,c,d,days=0;
    while(1)
    {
        days++;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        if (a+b+c+d == -4) break;
        for (i=d+1; ; i++) // pay attention: from d+1
        {
            if ((i-a)%23==0)
                if ((i-b)%28==0)
                    if ((i-c)%33==0)
                    {
                        printf("Case %d: the next triple peak occurs in %ld days.\n",days,i-d);
                        break;
                    }
        }
    }
    return 0;
}

Brian 2010-08-17 14:10 发表评论

POJ 1005 I Think I Need a Houseboat

Brian — Tue, 17 Aug 2010 06:09:00 GMT

Description

Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land, he learned that the state of Louisiana is actually shrinking by 50 square miles each year, due to erosion caused by the Mississippi River. Since Fred is hoping to live in this house the rest of his life, he needs to know if his land is going to be lost to erosion.

After doing more research, Fred has learned that the land that is being lost forms a semicircle. This semicircle is part of a circle centered at (0,0), with the line that bisects the circle being the X axis. Locations below the X axis are in the water. The semicircle has an area of 0 at the beginning of year 1. (Semicircle illustrated in the Figure.)

Input

The first line of input will be a positive integer indicating how many data sets will be included (N). Each of the next N lines will contain the X and Y Cartesian coordinates of the land Fred is considering. These will be floating point numbers measured in miles. The Y coordinate will be non-negative. (0,0) will not be given.

Output

For each data set, a single line of output should appear. This line should take the form of: “Property N: This property will begin eroding in year Z.” Where N is the data set (counting from 1), and Z is the first year (start from 1) this property will be within the semicircle AT THE END OF YEAR Z. Z must be an integer. After the last data set, this should print out “END OF OUTPUT.”

把题意弄明白，就知道这道题是水题了，由坐标 (0,0) 开始，以半圆为形状每年侵蚀50 平方 miles ，问你从 (0,0) 开始到 (x,y) 结束需要多长时间，水题不需要太关注效率，所以变量定义上没有深究，其他不解释。

编译器 C + + :
#include
using namespace std;
#define PI 3.1415926
int main()
{
     int n,i=0,year;
double x,y,area;
     cin>>n;
     while (i {
         cin>>x>>y;
         area = 0.5 * PI * (x*x+y*y); // semicircle area equation
         year = area/50;
         printf("Property %d: This property will begin eroding in year %d.\n",i+1,year+1);
   i++;
     }
     printf("END OF OUTPUT.\n");
return 0;
}

Brian 2010-08-17 14:09 发表评论

POJ 1004 Financial Management

Brian — Tue, 17 Aug 2010 06:06:00 GMT

正宗水题，就是输入12个浮点数，让你求平均值，不解释。

#include

int main()

{

float i=0,m,s=0;

for(;i<12;i++)

{

scanf("%f",&m);

s+=m;

}

printf("$%.2f\n",s/12);

return 0;

}

Brian 2010-08-17 14:06 发表评论

POJ 1003 Hangover

Brian — Tue, 17 Aug 2010 06:05:00 GMT

正宗水题，题目把最主要的公式都给你了，只要计算1/2+1/3+1/4+......+1/(n+1) >= x中最小的n值即可，我这里的cards用的是整形，注意底下一定要乘以1.0，否则会让你调试的生不如死的，要不你就让cards 是浮点型，其他的不解释。
#include

int main()
{
int cards;
float length,c;
for(scanf("%f",&c); c!=0.0; scanf("%f",&c))
{
  for(cards=0,length=0; length  {
   cards++;
   length+=1/(cards*1.0+1);
  }
  printf("%d card(s)\n",cards);
}
    return 0;
} //180K 0MS

Brian 2010-08-17 14:05 发表评论

POJ 1002 487-3279

Brian — Tue, 17 Aug 2010 06:04:00 GMT

事先申明，该程序虽然AC，但是效率极其低下，低下到让人发指的程度，我也不知道为什么。估计是用了STL的原因，具体我也说不清楚。其实思路不难，就是将字符转化成对应数字，然后将结果存放在一个整型向量中，接收字符串用的是字符串向量，处理的时候跟一般的字符串处理时一模一样的。处理结束之后要进行字典排序，显然要用排序函数，可以用冒泡，选择，快排，甚至是Hash，但是据说STL的sort 效率比快排还要快。源程序后附加了MSDN上的一些简单解释。没有翻译！

Description

Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You could order your pizza from Pizza Hut by calling their ``three tens'' number 3-10-10-10.

The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the mapping of letters to numbers, as follows:

A, B, and C map to 2
D, E, and F map to 3
G, H, and I map to 4
J, K, and L map to 5
M, N, and O map to 6
P, R, and S map to 7
T, U, and V map to 8
W, X, and Y map to 9

There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary. The standard form of TUT-GLOP is 888-4567, the standard form of 310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010.

Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)

Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more) businesses in the directory have the same telephone number.

Input

The input will consist of one case. The first line of the input specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters.

Output

Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line:
No duplicates.

C++ 编译器:

#include
#include
#include
#include // STL sort function
using namespace std;

char map[] = "2223334445556667#77888999#";
                  //ABCDEFGHIJKLMNOPQRSTUVWXYZ
void visited(char &ch) // visit and format strings
{
   if (ch >= 'A' && ch <= 'Z')
    ch=map[ch-'A']; // ch equals to its real number
}

int main()
{
    int N,i=0,j,flag=0;
string s;
vector stored(100000); // be visited & stored (up to 100,000)
    cin>>N;
vector counter(N,1); // stored times

for (; i {
  cin>>s;
  for (j=0; j  {
   visited(s[j]);
   if (s[j]!='-')
   {
    stored[i] += s[j];
    if (stored[i].length()==3)
     stored[i] += '-'; // 487 -[3] 3279
   }
  }
}
    sort(stored.begin(),stored.begin()+N); // Quicker than QuickSort!
// should not used stored.end() !
i=0; j=1;
    while (i {
  while(stored[i] == stored[j])
  {
   counter[i]++;
   j++;
   flag=1;
  }
  i=j;
  j++;
}

    if (flag)
  for (i=0; i  {
   if (counter[i]>1)
    cout<  } // must have { }
else cout<<"No duplicates."<
return 0;
}

Sort ：
Arranges the elements in a specified range into a nondescending order or according to an ordering criterion specified by a binary predicate.

template
   void sort(
      RandomAccessIterator _First,
      RandomAccessIterator _Last
   );

Brian 2010-08-17 14:04 发表评论

POJ 1001 Exponentiation

Brian — Tue, 17 Aug 2010 06:02:00 GMT

分类开篇语: 第一个程序搞了好几天，发现了很多问题。POJ不保证按顺序做且更新速度肯定不会很快。有些题自己做不出来借鉴别人的会注明出处。很多算法都需要从网上找，第一题的大浮点数相乘的核心算法就是这样找来的。我心里明白，虽然AC了，但是边缘数据处理的很粗糙，我自己都发现几个bug了，但是依然AC了。

本题主要注意将字符串转化为实际的数字然后借鉴数制的思想来进行大数相乘。

Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

Sample Input

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

编译器C++ 源码:

#include 
#include 
using namespace std;
#define MAX 255
  
int getnum(string s,int *c) // get real number of R
{ 
    int i=0,j=0,t[MAX];
    memset(t,0,sizeof(int)*MAX); // a stores 0
    
 while (i < 6) // R value 1 through 6
 {
  if (s[i] != '.')
  {
   t[j]=s[i]-'0';
   j++;
  }    
  i++;
 }     // a`s length = 5
    
 for (j=0; j<5; j++)
  c[j]=t[4-j]; // c stores in order from a
  
 for (i=0; s[i] != '.'; i++); // find decimal point
 return (5-i); // the position of . point
}
 
void multi(int *a,int *b) // big-multiplication
{ 
    int i=0,j,r=0,t[MAX];
    memset(t,0,sizeof(int)*MAX); // t stores 0
    
 for (; i<5; i++)
  for (j=0; j<255; j++)
   t[i+j] += a[i]*b[j]; // core algorithms!
 
 for (i=0; i<255; i++)
 {
  b[i]=(r+t[i])%10; // r always stores remainder
  r=(r+t[i])/10;   // b stores the result
 }     
}    // basic algorithms of b-m
int main() 
{ 
    int i,j,d_pos,n,a[MAX],b[MAX];
    string s;
    while (cin>>s>>n)
 {
  
  memset(b,0,sizeof(int)*MAX);
  memset(a,0,sizeof(int)*MAX);
  d_pos=getnum(s,a);
  getnum(s,b);
  
  for (i=0; i   multi(a,b);  // a is a loop invariant
  
  for (i=254; !b[i]; i--); //find last non-zero  
  for (j=0; !b[j]; j++); // find first non-zero
  
  for (; i >= n*d_pos; i--) // loop n times
   cout<  if (n*d_pos >= j+1) cout<<"."; //pay attention
  for (i=n*d_pos-1; i>=j; i--)
   cout<  
  cout< }
 return 0;
}

Brian 2010-08-17 14:02 发表评论

POJ 2379 震撼的打表!!!!!!

Brian — Tue, 17 Aug 2010 05:55:00 GMT

这不是解题报告，只是偶然看到的，以此与君共勉!

#include
unsigned short a[10002]={0,0,1,1,0,2,0,1,1,0,1,1,1,1,0,1,0,2,1,1,0,0,0,2,1,0,1,0,1,1,1,2,0,0,0,0,2,1,0,1,0,3,1,1,0,0,0,1,1,1,0,0,1,2,0,0,1,0,1,2,2,1,0,0,0,0,0,2,1,0,0,2,2,1,0,1,0,1,1,1,0,0,0,3,1,0,0,0,1,1,2,0,0,0,0,1,0,2,1,0,
2,2,1,1,0,0,0,1,0,2,0,0,2,1,0,0,0,0,0,2,2,1,0,0,1,0,0,2,1,1,0,2,1,0,0,0,0,1,2,2,0,0,0,2,1,0,0,0,0,1,1,1,2,0,0,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,0,1,2,0,0,0,0,0,1,2,2,0,0,1,0,1,2,0,0,0,1,1,1,0,1,0,3,1,3,
0,0,1,0,2,0,0,0,0,0,2,2,0,0,0,0,1,0,0,0,1,2,1,3,0,0,0,1,2,1,0,0,0,2,0,1,1,0,1,1,3,1,0,1,0,0,0,0,0,0,0,3,1,1,0,0,0,1,2,0,0,0,0,2,1,0,0,0,1,2,1,2,1,0,0,0,2,1,0,1,1,3,0,1,0,0,0,3,1,0,1,0,0,1,0,0,0,0,0,0,
2,1,0,0,2,0,1,1,1,0,0,5,0,1,0,0,0,1,1,1,1,0,0,2,1,0,1,0,1,1,2,2,0,0,0,0,0,1,0,0,3,1,0,0,0,0,0,1,1,2,0,1,1,2,0,0,0,0,1,1,1,0,0,1,1,0,0,1,0,0,1,3,2,2,0,0,1,0,0,2,0,1,1,1,2,0,0,0,0,1,2,0,0,0,0,2,1,1,0,1,
0,3,0,0,0,0,0,1,2,1,2,0,1,0,0,0,0,0,0,1,1,2,0,1,1,1,0,0,0,0,1,2,1,1,2,1,0,0,1,3,1,0,1,2,0,0,0,0,0,2,1,0,0,0,0,0,2,2,0,0,1,1,2,2,0,0,0,1,1,0,0,1,1,2,1,0,0,0,0,2,2,0,0,0,1,0,0,2,0,0,0,3,2,1,0,0,1,1,0,2,
0,1,0,2,0,0,0,0,2,1,3,0,0,0,0,1,0,1,0,1,1,1,1,2,0,0,0,1,0,0,0,0,1,2,1,0,0,0,0,1,2,1,0,0,0,0,1,1,1,0,0,2,2,1,0,0,1,1,2,2,0,0,1,2,1,1,1,0,1,1,0,1,0,0,0,0,2,1,0,0,1,2,1,1,0,0,0,2,0,1,0,0,0,3,1,0,1,0,1,1,
1,1,0,0,0,0,1,2,0,0,0,1,1,1,0,0,0,2,1,1,2,0,1,0,1,1,0,0,0,1,2,2,0,1,1,2,1,1,1,1,0,1,0,1,0,0,0,2,1,0,0,0,0,1,1,0,0,0,0,2,3,2,1,0,0,0,0,0,1,0,0,1,0,1,0,0,0,2,0,2,0,0,2,2,2,0,0,0,0,2,1,2,0,0,0,1,2,0,0,0,
1,2,2,0,0,0,0,2,0,1,1,0,2,1,2,0,0,0,0,2,0,1,0,0,2,1,1,1,0,0,0,1,2,3,0,0,0,0,0,1,1,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,2,0,1,0,1,2,2,0,0,1,0,1,1,0,1,1,1,2,0,0,0,0,1,2,1,0,0,0,0,2,2,0,2,0,2,0,0,0,0,1,2,1,0,
1,0,1,3,0,0,0,0,0,1,1,2,0,0,0,0,1,1,0,0,0,1,1,1,1,0,0,2,1,2,1,0,0,2,2,0,0,0,0,2,1,2,0,0,0,0,2,0,0,0,0,0,1,1,0,0,0,3,0,1,1,0,0,5,3,0,0,0,0,1,1,0,2,0,1,0,0,1,0,0,1,2,1,3,0,0,1,1,1,0,1,1,1,0,0,0,1,0,0,0,
1,0,0,0,0,2,1,1,0,0,0,2,2,0,0,0,0,1,1,1,0,0,1,1,1,0,1,0,2,2,1,3,0,0,0,1,1,1,0,0,0,3,1,0,0,0,2,2,0,0,0,0,0,2,1,0,0,0,1,0,2,3,0,1,2,0,2,1,0,0,0,1,0,0,0,0,0,2,1,1,0,0,0,3,0,1,0,0,1,0,3,3,0,0,0,0,0,1,0,0,
0,1,2,0,0,0,0,0,1,1,0,0,3,1,0,1,0,0,0,2,1,1,0,1,0,2,1,1,0,0,1,1,2,2,0,0,0,0,0,1,2,0,0,2,1,0,0,0,0,2,1,1,0,0,1,1,1,1,1,0,3,3,0,2,1,0,0,2,0,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,2,0,0,1,1,1,1,1,1,1,0,0,0,1,1,3,
0,0,0,2,3,0,0,0,0,1,0,0,1,0,2,1,0,1,0,1,2,0,0,2,0,0,0,0,0,1,0,0,1,2,1,2,0,0,0,2,2,0,0,0,1,0,0,0,1,0,0,4,2,2,1,0,1,0,0,1,0,1,0,2,4,0,0,0,0,2,0,2,1,0,1,0,0,0,0,0,1,1,0,0,0,0,0,3,0,0,0,0,1,2,2,0,0,1,0,1,
1,1,0,0,0,0,1,1,1,0,0,2,2,2,1,0,1,1,0,1,1,1,0,2,1,0,0,0,0,2,1,1,0,0,0,0,3,1,0,0,0,0,2,0,0,0,0,2,1,2,2,1,0,1,1,0,0,0,0,2,0,0,1,0,1,1,1,1,0,0,0,2,2,0,0,0,1,1,1,1,1,2,0,3,1,1,1,0,0,1,1,1,0,0,1,0,0,1,1,0,
1,1,0,2,0,0,0,1,0,0,0,1,2,3,0,1,0,0,0,1,4,2,1,0,0,0,0,1,0,0,1,1,1,0,1,0,1,1,0,1,0,0,0,1,0,0,0,0,0,2,2,2,0,0,1,1,0,0,0,0,1,3,2,0,1,0,1,4,0,1,0,1,0,1,1,0,1,0,0,0,1,3,0,0,0,0,1,0,0,0,0,1,3,0,0,1,1,0,0,1,
0,0,1,1,1,1,0,0,1,2,1,1,0,0,0,2,0,0,0,0,2,0,1,2,0,0,0,1,2,1,0,0,0,3,1,0,0,0,0,3,0,2,0,0,1,0,1,1,0,0,1,1,2,1,0,0,0,1,0,2,1,0,0,1,2,0,0,0,0,0,2,1,1,0,0,2,0,1,2,1,2,1,2,2,1,1,0,2,1,1,0,0,0,3,2,0,0,0,0,1,
0,0,1,0,0,0,0,1,1,0,0,3,0,0,0,0,1,0,1,1,1,0,0,3,3,0,0,0,0,0,1,1,0,1,0,1,0,0,0,0,0,1,1,2,0,0,1,0,2,1,0,0,2,3,0,1,0,0,0,2,2,1,0,0,1,1,0,1,0,0,2,1,0,0,1,0,1,0,0,2,0,0,0,2,1,1,1,0,2,0,0,2,0,1,0,0,2,1,0,0,
0,2,0,1,0,0,1,3,0,2,0,1,0,2,0,0,0,0,1,1,2,2,1,0,0,0,2,1,0,0,0,0,1,0,0,0,1,2,0,0,0,0,0,3,2,2,0,0,0,0,4,0,0,0,1,0,1,1,0,0,1,0,1,1,0,0,0,2,1,1,1,0,0,1,1,0,0,1,0,1,1,1,0,0,0,0,0,2,0,1,0,2,2,1,0,0,0,1,0,1,
0,0,0,2,2,0,0,1,1,1,3,0,2,1,0,2,2,2,1,0,1,1,2,1,1,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,1,1,0,0,0,1,2,2,0,0,1,1,2,3,0,0,0,1,2,1,0,0,0,0,2,0,0,0,0,0,1,0,0,0,1,2,0,1,0,1,0,0,0,1,1,0,1,2,3,1,0,2,0,3,1,0,0,0,0,1,
1,3,1,0,0,0,0,1,0,0,2,1,0,0,0,0,0,0,1,0,0,1,0,3,1,0,0,0,1,1,2,2,0,0,1,0,1,1,0,0,0,1,0,1,0,0,2,1,2,2,0,2,0,0,4,0,0,0,0,2,0,2,0,0,0,0,1,2,0,1,1,1,2,2,0,0,0,1,1,1,0,0,0,0,0,0,0,0,2,2,0,2,0,0,0,2,1,0,0,0,
2,3,0,1,0,0,0,1,0,0,0,1,0,1,1,1,1,0,1,0,1,1,0,0,1,0,2,0,0,0,2,2,0,1,0,0,0,1,1,0,1,0,0,1,0,1,0,1,2,2,0,4,0,0,0,0,3,0,0,1,2,0,0,0,0,0,0,2,0,0,0,1,0,2,2,0,0,0,1,2,2,1,1,1,0,0,1,2,2,0,0,0,0,2,1,1,0,1,1,2,
0,0,2,1,0,0,1,0,0,0,0,3,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,3,0,1,1,0,2,1,1,1,0,0,1,1,3,0,0,0,0,1,1,1,0,0,1,1,1,2,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,2,0,0,1,0,0,1,2,1,0,1,0,2,0,2,1,0,1,1,1,1,0,0,0,0,1,0,0,0,0,3,
2,1,0,0,0,2,0,0,1,0,2,3,2,2,0,0,1,1,0,0,0,0,1,1,2,1,0,1,0,2,0,1,2,0,0,0,0,1,0,0,0,2,1,2,1,0,1,1,1,0,0,0,1,1,0,2,1,0,1,0,0,1,1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,1,1,3,1,0,0,1,1,0,0,0,0,1,1,0,1,0,1,1,0,0,0,2,
1,0,1,3,0,0,1,1,1,2,0,0,1,2,1,1,0,0,1,0,0,2,0,0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,2,1,0,0,1,1,0,1,0,1,1,2,2,1,1,0,0,1,0,0,1,1,0,0,2,0,0,1,3,0,2,0,1,0,2,1,1,1,0,0,0,2,1,0,0,0,0,1,2,1,0,0,0,1,1,0,0,1,2,1,0,
0,1,0,3,3,1,0,1,1,1,1,2,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,2,1,1,0,0,0,1,1,0,1,0,0,1,0,1,0,0,0,0,2,2,0,1,1,2,1,0,0,0,1,3,1,0,0,0,0,0,0,0,1,0,1,1,1,1,0,0,2,1,1,1,0,0,2,1,1,2,0,1,1,3,0,1,0,0,1,4,1,0,0,0,1,2,
1,0,0,0,0,0,0,0,0,0,1,1,0,1,1,0,0,3,0,1,1,0,1,2,1,1,0,1,0,0,1,0,1,0,0,1,3,1,1,0,1,2,0,2,0,1,0,1,1,0,1,0,1,0,0,2,1,0,0,2,2,1,0,0,0,0,1,1,1,0,0,1,0,1,1,1,0,2,1,1,2,0,0,0,2,1,1,0,0,1,0,0,1,0,0,1,0,0,0,0,
0,1,1,1,0,0,0,3,2,1,0,0,0,0,1,0,0,0,0,1,0,2,0,1,0,0,1,1,0,0,1,1,0,2,2,0,1,0,0,1,0,0,0,2,0,0,0,0,1,2,1,1,1,0,1,2,3,1,0,0,0,0,2,1,1,0,0,2,0,1,0,0,1,2,1,0,0,0,1,3,1,0,2,0,2,0,3,2,1,0,0,1,0,1,0,0,0,1,1,1,
0,0,1,2,1,1,0,0,0,1,1,0,2,0,0,0,1,1,1,0,1,2,0,0,0,0,1,1,0,0,0,0,1,2,2,0,0,0,0,0,1,2,0,1,0,0,1,4,1,0,0,1,1,1,0,0,1,2,1,1,0,0,0,1,0,0,0,0,0,0,3,1,0,1,0,2,1,2,0,0,0,0,1,1,0,0,1,1,1,4,1,0,0,1,1,0,0,0,1,1,
1,0,0,1,0,1,3,1,0,0,0,1,1,2,0,0,0,1,1,2,0,0,0,0,1,0,0,0,1,1,1,3,0,0,0,1,0,2,0,0,2,1,1,1,0,1,0,2,0,1,0,0,2,1,2,0,0,1,0,2,1,1,0,0,2,1,1,1,0,0,1,1,0,1,0,0,1,3,0,0,1,0,0,0,1,2,0,0,0,1,0,1,0,0,0,0,0,1,0,0,
0,1,0,1,0,0,1,2,2,0,1,1,1,0,2,1,1,0,0,2,1,0,0,0,0,1,1,0,0,0,0,0,1,1,0,0,0,3,0,1,0,2,2,1,0,1,0,1,1,1,1,1,0,0,0,0,4,1,0,0,0,1,1,0,1,1,0,4,2,0,0,0,1,1,0,0,0,0,0,2,0,0,0,0,0,1,3,1,1,0,1,2,0,0,0,0,0,3,1,0,
1,0,0,1,2,1,0,0,0,2,1,1,2,0,2,1,1,1,1,0,0,0,1,1,0,1,2,3,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1,0,0,0,0,1,2,2,0,0,0,3,2,0,0,0,0,3,1,1,1,0,2,1,1,2,0,0,0,0,1,0,0,0,1,0,1,0,1,1,0,2,0,0,2,0,1,1,1,1,0,1,0,2,
1,1,0,0,0,1,0,1,0,0,3,1,1,1,0,0,0,1,0,2,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,1,0,3,2,1,1,2,0,0,0,0,2,0,1,1,1,0,0,1,1,0,0,1,0,1,1,1,0,1,0,1,2,2,1,1,2,1,1,2,2,0,0,1,0,1,0,0,1,2,0,2,0,2,0,1,1,0,1,0,1,0,1,0,1,0,
0,0,1,1,0,0,1,0,0,1,0,0,1,2,0,0,0,0,0,3,1,3,0,0,0,1,2,0,0,1,0,0,0,1,0,0,0,3,1,1,0,0,0,0,1,1,0,0,0,0,2,1,0,0,2,0,0,0,1,0,0,1,2,1,0,0,3,3,1,1,0,0,0,0,1,1,0,0,0,1,1,3,0,0,0,1,0,1,1,1,0,2,1,1,0,1,0,0,4,0,
0,0,0,1,1,0,0,0,2,2,0,0,0,0,0,1,1,1,0,2,0,1,1,1,0,1,0,0,0,1,0,0,3,0,0,0,0,1,2,0,1,0,0,1,0,0,1,2,1,1,2,2,1,2,0,0,1,2,1,1,0,2,0,3,4,0,2,0,0,0,0,2,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,0,0,1,0,0,2,1,1,0,0,0,2,
3,1,0,0,1,1,0,1,0,1,2,0,0,2,0,0,1,0,2,1,1,2,0,2,1,0,0,0,0,2,2,1,0,0,1,1,1,0,0,0,0,0,0,1,0,0,0,3,2,2,1,0,0,2,0,0,0,1,0,1,0,1,2,0,0,0,0,0,0,1,1,1,0,1,0,0,1,0,2,0,0,0,0,0,2,1,0,1,0,4,2,2,0,0,0,0,2,3,0,0,
1,1,0,2,0,0,1,1,3,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,2,1,0,0,0,0,2,0,1,2,0,1,1,1,0,0,0,1,2,0,2,1,1,1,0,1,1,0,0,0,0,0,2,2,0,0,2,2,1,1,0,1,0,2,0,2,0,0,1,0,1,1,0,0,1,0,1,0,0,0,0,0,0,1,1,1,0,1,1,0,0,0,1,1,0,2,
2,0,0,3,1,0,0,0,0,1,0,1,2,0,1,2,1,3,0,0,0,0,0,0,1,0,0,2,2,1,0,0,0,2,1,0,1,0,1,2,0,2,0,1,1,0,0,2,0,0,1,0,0,0,1,0,0,1,1,1,1,2,0,0,2,0,0,0,0,0,1,2,0,0,0,0,2,0,1,0,1,2,0,1,0,0,0,2,3,1,2,1,0,2,0,0,0,1,0,0,
1,2,0,0,1,0,0,2,0,0,1,0,2,1,1,0,0,3,0,0,0,1,0,1,2,0,0,1,1,1,2,2,0,1,2,0,2,2,1,0,0,0,0,2,0,0,0,2,0,0,0,0,0,1,2,0,1,0,1,2,0,1,0,0,0,0,1,0,1,1,0,3,0,1,0,0,0,1,1,0,0,0,0,2,0,0,0,1,1,0,0,2,0,2,2,0,2,2,1,0,
1,2,1,1,0,0,1,0,1,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,3,2,2,0,0,1,0,1,0,0,0,0,2,2,0,0,0,0,1,1,2,1,0,0,2,1,0,0,4,1,1,2,0,0,0,2,0,0,1,0,0,1,1,1,0,0,0,2,0,0,1,0,1,1,1,1,0,1,0,1,0,2,0,0,3,0,0,2,0,0,0,1,1,1,
0,0,0,2,1,1,0,0,2,0,1,1,0,0,1,1,0,0,0,0,1,3,1,1,0,1,1,0,1,1,0,2,0,1,1,0,0,1,1,1,0,0,1,0,1,1,1,2,0,0,1,2,0,3,1,0,1,0,0,0,0,0,0,2,1,0,0,0,3,1,3,1,0,0,1,1,1,2,0,0,0,1,1,0,0,0,0,1,0,2,0,0,0,0,0,1,0,0,2,2,
2,2,0,0,1,0,0,1,0,0,0,3,0,0,0,0,0,1,0,1,0,1,1,1,1,2,0,0,1,2,0,1,0,0,1,1,1,2,1,0,0,0,0,1,0,0,1,2,1,1,0,1,1,1,2,0,3,0,0,0,0,0,0,0,0,0,1,3,0,0,2,2,0,0,0,0,0,1,0,2,2,0,0,0,1,0,0,1,0,2,1,0,0,0,0,0,2,1,1,0,
1,1,2,1,0,1,1,1,0,1,1,1,1,1,2,0,0,3,1,1,0,1,1,1,1,1,1,1,2,0,1,0,0,1,0,0,0,0,1,0,0,0,0,1,2,0,0,0,1,1,0,1,0,0,1,1,1,3,0,1,0,2,3,0,1,0,0,1,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,1,0,2,0,0,1,0,3,1,3,0,0,0,3,1,2,
0,1,0,0,1,0,0,0,0,1,3,2,0,0,0,0,1,0,0,0,0,1,1,1,1,0,0,2,0,1,0,0,1,1,2,1,1,0,0,2,0,0,0,0,1,1,0,0,0,0,2,1,2,2,1,0,0,2,0,2,1,0,0,0,1,0,0,0,0,1,1,1,1,1,1,0,2,2,0,0,1,1,1,0,0,0,1,0,1,2,0,0,0,0,0,0,0,1,0,1,
4,1,0,0,0,1,0,0,1,0,2,2,0,0,0,0,0,3,1,1,0,1,1,0,2,1,0,2,0,2,2,2,1,0,0,0,0,2,0,0,0,2,2,2,1,0,0,1,0,2,0,0,1,2,0,0,0,0,0,2,1,1,0,0,1,1,0,1,1,0,0,2,0,1,0,0,0,0,2,1,3,0,0,1,2,0,2,1,0,2,1,0,0,0,0,0,1,2,0,0,
0,1,1,0,0,0,0,1,0,2,1,0,0,0,1,0,0,0,1,1,1,0,0,0,0,2,1,1,1,1,1,1,0,1,0,0,0,3,1,1,2,0,0,0,1,0,0,0,0,1,2,1,0,0,0,1,0,2,0,1,0,3,1,3,1,0,0,0,1,1,2,1,0,1,0,0,0,0,0,0,2,0,1,1,1,0,0,2,0,0,0,2,1,0,1,0,0,2,0,0,
0,0,1,0,0,1,1,0,0,1,2,2,1,0,0,0,1,0,0,1,1,2,0,2,0,0,1,0,2,0,0,0,0,2,2,3,1,0,1,0,0,1,0,2,1,0,0,1,2,0,1,3,1,0,0,1,0,2,2,2,0,0,0,1,2,0,0,0,0,2,0,0,0,0,1,0,2,0,2,1,0,2,1,1,1,0,0,1,1,0,0,0,0,2,1,0,0,0,0,0,
0,2,0,1,2,0,1,3,0,0,0,0,1,1,0,0,0,1,1,2,2,0,1,1,0,0,0,0,0,1,0,0,1,1,0,2,1,0,0,0,0,0,1,0,0,1,0,1,1,3,0,0,0,2,3,0,1,1,0,1,1,2,0,0,1,2,1,1,1,0,0,0,0,0,0,0,0,1,2,0,1,0,0,1,1,0,0,0,0,0,1,1,0,0,1,0,1,3,1,0,
0,1,1,3,0,0,1,1,0,0,0,0,1,2,0,0,2,0,1,0,2,2,0,1,0,0,1,0,0,0,0,1,0,1,0,1,1,3,2,2,1,0,0,1,3,1,0,0,0,1,2,2,0,1,0,0,1,1,0,1,1,2,0,1,0,0,1,0,1,1,0,1,1,2,0,0,0,1,0,2,1,1,0,0,0,0,0,0,1,0,0,3,0,0,0,0,2,2,1,1,
0,0,1,1,1,0,0,1,2,0,1,1,0,0,1,1,1,1,0,0,0,1,0,1,0,0,0,0,3,2,0,0,1,2,0,0,2,1,0,0,0,0,0,1,0,3,1,0,2,0,1,2,0,0,0,0,1,0,1,3,0,0,1,2,1,1,0,0,1,0,1,0,1,0,0,0,0,0,0,1,0,1,2,1,0,1,0,1,1,2,0,0,1,1,2,1,0,0,2,1,
0,1,1,0,0,1,1,0,0,0,1,1,0,1,0,1,1,3,0,0,0,0,0,0,1,2,1,0,2,1,1,1,0,1,1,0,2,0,0,0,1,2,0,1,0,0,0,1,1,1,0,1,0,2,1,0,0,0,1,0,2,1,1,0,0,2,0,2,0,0,0,2,0,0,1,0,0,2,2,1,2,1,2,2,0,1,1,0,3,2,0,0,0,0,0,1,0,0,0,0,
1,1,1,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,2,4,2,0,0,0,1,1,0,1,0,0,1,0,1,0,0,0,1,0,0,1,0,1,2,2,0,1,0,0,1,3,1,1,0,0,0,1,4,0,0,0,0,0,1,0,0,1,2,0,3,0,0,1,1,0,0,1,0,0,1,1,0,0,0,0,0,0,2,1,0,0,1,1,2,1,0,0,0,1,3,
0,0,0,1,0,0,0,0,1,3,0,1,0,0,2,0,0,1,1,3,1,3,2,1,1,0,0,0,1,2,0,0,0,0,0,1,0,0,0,2,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,1,0,4,0,0,1,0,3,0,0,0,1,2,1,1,0,0,2,0,1,1,0,0,0,2,2,0,0,1,0,1,2,2,0,0,1,1,0,0,1,0,1,1,
4,1,0,0,0,2,0,2,1,0,1,0,2,1,0,2,0,2,2,2,0,2,0,0,0,0,0,0,1,1,0,0,0,0,0,2,1,0,0,0,0,0,2,1,0,0,0,1,1,1,1,0,0,4,1,0,0,0,0,1,2,1,0,1,0,0,0,1,0,1,1,1,0,1,1,0,0,2,3,1,0,0,0,0,1,0,0,0,2,1,2,1,1,1,0,0,0,2,1,1,
0,0,1,1,0,0,0,1,3,1,0,1,0,0,0,1,0,1,0,0,1,2,0,0,0,0,2,1,0,0,2,1,0,2,0,1,0,1,1,0,0,2,0,0,0,0,0,0,0,0,3,0,0,1,3,1,0,0,0,1,0,3,2,1,0,1,0,1,0,0,0,0,1,2,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,2,1,2,2,0,0,2,1,2,0,
1,0,0,2,1,1,0,0,1,2,0,0,0,0,0,0,2,2,0,1,0,0,2,1,0,0,0,0,1,1,0,1,1,2,1,0,0,2,0,0,1,1,0,0,1,1,2,2,2,0,1,2,0,0,0,0,0,0,0,1,1,1,1,0,0,2,0,0,0,1,1,1,0,0,0,2,2,1,0,0,0,1,1,0,0,0,0,1,1,0,2,0,2,2,0,2,0,0,2,1,
0,1,1,2,1,0,1,1,0,0,1,0,0,2,0,0,1,1,1,1,0,2,0,1,1,0,0,0,1,1,3,1,1,0,0,1,1,2,0,0,1,1,1,1,0,0,1,0,1,2,0,0,0,1,1,0,0,0,0,0,2,0,0,0,1,1,0,1,0,1,0,3,1,0,0,0,0,1,1,1,0,0,1,2,0,0,0,0,0,1,2,1,0,1,0,2,0,0,0,1,
1,2,2,1,0,0,0,2,1,0,0,0,1,1,2,0,0,0,1,2,2,2,0,0,0,0,0,1,1,0,1,2,0,0,0,0,0,1,0,1,0,1,0,0,2,0,0,1,2,3,0,0,0,0,0,1,1,1,0,0,0,1,1,1,0,0,1,3,2,1,0,1,2,1,0,1,0,1,0,1,3,2,0,0,2,0,0,1,1,1,0,1,1,1,0,0,0,0,1,1,
1,0,0,0,1,0,1,0,0,0,0,1,2,0,0,0,0,0,0,1,0,2,1,2,0,0,2,2,2,2,0,0,0,1,0,0,0,0,2,1,1,2,0,0,0,1,0,1,0,0,1,2,1,1,0,0,0,2,1,1,0,1,0,0,0,0,0,0,1,2,2,1,0,1,1,1,0,0,0,0,1,0,0,3,1,0,0,0,1,1,0,1,1,2,1,0,0,1,1,1,
1,2,1,0,1,0,2,0,2,1,0,2,1,0,0,0,1,2,1,1,0,1,0,1,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,1,2,0,1,0,0,0,0,1,1,0,1,0,3,0,0,1,0,0,2,0,0,1,0,1,0,2,1,0,1,4,2,1,0,0,0,0,2,0,1,0,0,0,1,1,0,1,0,0,1,1,2,0,0,0,0,0,0,1,0,
2,1,1,1,1,0,1,1,0,0,3,0,1,3,0,0,0,0,0,1,1,1,1,1,1,2,2,2,1,0,1,0,0,1,1,1,0,1,0,1,0,0,1,2,1,0,0,0,3,3,1,1,0,0,0,2,0,1,0,0,1,1,0,1,1,0,1,1,0,1,0,0,0,1,0,0,0,2,1,1,1,1,0,0,1,3,1,1,0,1,1,0,1,0,1,1,0,1,1,0,
0,0,0,3,0,1,0,1,0,0,2,0,0,0,0,0,0,0,0,0,1,2,1,1,0,0,0,2,1,0,1,0,2,1,1,0,0,1,1,2,3,0,0,0,1,0,0,0,0,0,0,0,2,2,0,0,0,1,0,1,1,0,1,0,1,0,0,0,1,0,2,0,0,0,0,1,0,2,0,2,2,3,0,0,1,0,0,2,0,1,0,0,1,0,1,0,1,0,1,4,
1,1,0,0,1,0,0,1,0,0,2,2,3,2,1,0,0,0,0,0,0,0,0,2,0,0,2,0,0,2,2,1,0,1,0,0,0,1,0,0,1,0,1,2,1,1,0,3,0,0,1,0,0,2,0,1,0,1,0,0,1,0,0,0,2,1,0,1,0,0,0,2,1,3,0,0,2,0,1,3,1,2,0,1,0,0,0,0,0,1,3,1,0,0,0,1,0,0,0,0,
3,1,0,0,0,0,1,2,2,0,1,0,0,2,1,1,0,1,0,1,0,2,0,1,1,0,1,0,1,0,0,2,0,2,0,0,1,1,1,0,0,1,0,1,0,0,3,1,1,1,0,1,0,0,1,2,1,1,1,0,0,1,1,2,1,0,0,0,0,1,0,0,2,2,0,0,1,0,2,0,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,0,0,1,1,1,
0,0,0,1,2,1,0,0,0,2,1,1,0,1,0,0,1,1,1,1,2,1,0,2,1,0,1,1,2,2,0,2,0,1,0,0,0,0,1,0,1,1,0,0,0,0,0,1,0,1,1,1,0,1,0,0,0,3,3,1,0,1,1,1,2,1,0,1,0,1,0,1,1,1,0,1,1,1,0,0,1,0,0,0,0,0,0,2,0,0,0,0,1,0,0,0,1,1,0,2,
2,2,2,0,0,1,2,0,0,2,2,3,0,1,0,0,0,2,0,0,1,1,0,2,0,0,0,0,3,1,3,1,1,1,0,0,2,1,1,0,1,0,0,1,0,0,0,1,0,1,2,0,1,2,0,0,0,0,1,1,2,1,0,1,0,0,0,1,1,1,0,1,0,3,0,0,0,1,1,2,2,0,0,0,1,0,0,1,1,2,1,0,0,0,2,1,1,2,0,0,
0,1,0,0,1,0,0,1,0,1,0,0,1,0,1,0,0,0,0,1,1,2,2,0,1,0,1,1,0,1,2,1,0,0,0,1,1,2,1,1,0,1,0,0,1,2,0,0,2,2,1,1,0,0,0,2,0,1,0,1,0,1,1,0,1,0,1,0,0,2,0,0,1,1,0,0,0,0,0,1,3,2,1,0,0,0,0,0,0,1,0,3,1,1,0,0,0,1,0,1,
2,1,3,0,4,0,0,0,0,1,2,0,0,1,0,0,1,0,0,0,0,1,0,0,1,0,0,1,0,1,1,0,0,0,1,0,0,1,1,0,1,0,2,2,1,0,0,2,0,0,0,2,0,3,0,1,1,0,0,0,1,0,0,1,0,0,0,0,2,2,1,1,0,2,0,0,0,2,0,2,2,1,0,0,2,0,0,1,1,3,1,1,0,1,1,0,1,0,0,2,
1,1,0,0,0,0,0,1,1,1,1,1,1,0,0,1,1,1,0,1,2,0,1,1,1,0,0,1,1,0,0,0,2,0,0,1,0,2,0,2,1,1,2,0,1,1,0,0,0,1,0,0,1,1,0,0,1,0,1,2,3,1,0,0,0,0,0,0,1,0,0,1,0,2,1,0,0,0,1,1,0,0,1,0,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,
0,2,0,1,0,0,2,1,1,1,1,1,0,0,0,1,0,0,1,2,1,0,0,0,0,0,1,0,0,0,0,1,2,2,0,0,1,1,0,1,0,1,0,0,2,1,1,0,1,2,2,1,1,1,1,0,0,0,1,0,1,2,2,2,0,0,1,1,0,0,1,0,0,0,1,0,1,0,0,3,1,1,1,0,0,0,0,1,0,2,0,1,0,2,0,1,0,0,2,0,
0,0,1,1,1,0,0,0,0,2,0,1,1,2,0,0,0,0,0,0,1,1,0,2,0,1,2,1,1,2,0,0,0,1,0,0,1,0,0,2,0,3,1,0,0,1,1,0,0,1,2,0,1,1,0,0,0,1,1,1,2,1,2,1,0,1,0,0,1,1,2,2,0,0,0,0,0,2,1,0,0,2,1,1,0,0,1,0,1,1,1,0,0,0,0,1,0,0,0,2,
2,0,0,1,1,1,1,1,1,0,0,1,2,0,0,0,0,3,2,0,0,0,1,1,1,1,0,0,0,1,2,1,0,0,0,0,1,0,1,1,1,1,0,0,2,0,0,2,2,2,0,0,0,1,0,1,0,0,1,1,2,1,0,0,0,0,0,1,0,0,0,2,0,0,2,0,0,1,1,0,0,1,0,2,1,1,1,1,0,3,1,2,0,0,0,0,0,1,1,0,
1,1,0,0,0,0,1,0,0,0,1,2,1,3,1,1,0,1,0,2,1,0,0,0,0,0,0,2,1,0,0,0,0,1,0,0,1,1,0,3,0,1,1,1,1,0,0,0,2,0,0,1,1,0,1,1,3,2,0,0,1,0,1,0,1,0,0,1,1,1,0,0,2,0,0,1,0,0,1,3,1,1,0,1,0,1,0,1,1,0,1,1,1,0,0,0,2,0,0,0,
1,1,1,1,2,1,1,1,1,1,0,0,0,1,1,0,2,0,0,0,0,3,1,1,1,0,0,1,1,3,1,2,0,0,1,0,2,1,0,1,0,2,0,0,0,0,1,2,0,0,2,3,2,1,0,0,0,0,0,1,1,2,0,0,1,2,0,0,0,0,0,0,0,1,0,1,1,1,0,0,1,1,0,0,1,0,0,2,1,0,0,2,1,1,2,1,1,0,1,1,
1,0,0,1,0,1,0,1,0,0,0,2,1,1,0,1,0,1,1,1,1,1,1,1,1,0,0,1,0,1,1,0,0,0,1,0,0,1,0,0,1,1,0,2,0,0,0,2,1,2,0,1,0,1,2,0,2,1,1,2,0,0,1,0,0,0,1,0,1,0,3,1,1,0,0,0,1,3,0,1,1,1,0,1,0,1,1,0,0,1,0,0,0,1,0,0,0,1,0,0,
1,1,2,0,0,1,0,1,0,1,0,0,2,0,2,1,0,1,3,0,0,1,0,0,0,1,0,1,1,1,1,1,0,1,0,0,0,0,0,0,1,2,1,2,1,0,0,2,0,2,2,1,0,0,0,1,0,0,0,0,2,0,1,0,0,1,0,0,2,1,0,1,1,1,0,0,0,1,1,0,0,1,0,0,0,0,2,0,1,0,1,0,0,1,0,0,1,2,2,1,
0,2,0,0,0,0,1,0,2,3,2,2,0,2,1,0,1,3,3,1,0,1,0,0,0,0,0,1,2,1,0,0,1,2,0,1,0,0,0,0,1,0,0,0,2,1,2,0,0,0,0,4,0,1,0,0,0,1,1,1,1,1,0,0,1,0,0,0,0,1,0,1,3,1,0,0,0,1,0,1,0,2,1,0,1,0,1,1,1,1,0,0,0,0,0,0,0,0,0,1,
2,1,1,0,1,1,0,1,1,0,0,0,1,0,1,2,0,1,1,1,0,0,2,1,0,1,0,0,2,1,1,0,2,0,0,1,1,1,0,1,1,2,1,0,0,0,1,2,1,2,0,0,0,1,0,2,1,1,1,1,0,2,0,1,1,0,0,0,0,0,1,2,1,1,0,0,1,1,1,0,0,0,0,1,1,0,0,0,0,1,1,2,1,0,0,0,0,0,0,1,
1,3,1,2,0,0,0,3,0,0,2,1,0,2,1,0,0,0,0,1,0,2,2,0,3,0,0,1,0,1,0,1,0,1,1,0,2,0,0,1,0,1,0,2,1,0,0,0,0,3,0,1,1,0,1,0,1,2,1,0,2,0,0,0,1,0,0,0,0,1,0,1,1,2,0,0,1,0,2,0,1,1,1,0,1,1,1,3,0,1,0,1,0,1,1,0,0,2,2,3,
0,1,0,1,1,0,0,1,1,0,0,0,0,1,0,0,2,1,0,1,1,0,0,1,0,0,2,3,0,0,1,0,2,1,1,1,0,0,0,1,2,2,0,0,1,0,0,1,0,0,0,2,0,1,0,0,0,1,2,1,2,0,0,2,3,0,0,0,1,0,1,2,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,0,2,0,3,0,1,1,0,0,2,
1,0,0,1,1,0,2,0,0,0,0,0,0,0,0,0,0,1,2,2,0,0,1,1,0,1,0,2,2,3,1,0,0,1,0,2,2,0,1,0,1,1,1,0,0,0,0,1,2,0,0,1,2,2,0,0,0,0,1,1,1,0,0,0,0,0,1,2,0,0,1,1,0,3,1,0,0,1,3,1,0,0,1,1,0,1,0,0,0,0,2,0,0,0,1,0,0,1,0,1,
0,4,0,0,0,1,0,1,2,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,1,0,1,0,0,2,1,1,1,0,0,1,3,2,0,0,1,1,1,0,1,0,2,0,1,1,1,0,2,3,0,3,0,0,0,2,1,0,2,0,1,0,1,0,0,0,0,1,1,1,2,1,1,1,1,1,1,1,0,1,0,1,1,0,0,1,0,1,2,1,0,0,1,1,0,
0,0,1,1,1,0,0,1,0,1,3,2,0,0,1,0,0,1,0,1,1,0,0,0,0,0,0,1,0,0,1,2,1,0,0,0,0,1,1,1,1,0,0,1,0,2,2,1,2,1,0,1,0,1,0,1,0,3,1,2,0,0,0,0,2,0,0,0,2,3,0,1,0,0,0,0,1,1,1,0,0,3,0,0,1,0,0,1,0,2,0,0,0,2,0,0,1,0,0,1,
1,1,1,0,0,0,0,0,1,1,0,2,3,0,1,1,2,3,0,1,0,0,0,2,0,0,1,0,0,0,2,2,0,0,1,2,0,0,0,0,2,0,2,1,0,0,2,1,1,0,0,1,4,1,0,0,0,0,0,0,0,3,1,0,0,1,0,1,0,0,1,1,0,0,0,1,0,0,0,2,1,2,0,0,2,0,0,0,0,0,0,2,1,0,0,1,1,0,1,1,
1,2,0,1,0,1,0,2,1,2,1,0,0,1,1,0,0,1,0,1,1,2,0,0,1,0,1,0,0,0,0,1,0,1,1,0,0,1,1,0,0,0,0,3,0,0,1,1,0,1,0,1,0,0,2,0,2,0,2,3,1,2,2,1,0,2,2,0,0,3,1,0,1,2,0,1,0,0,0,1,1,0,0,0,0,0,1,1,0,0,2,1,1,2,0,0,1,2,1,0,
0,1,0,1,2,0,0,0,0,1,1,2,0,0,0,0,1,1,0,1,1,0,0,0,0,1,1,0,2,1,0,1,1,2,0,1,1,1,0,0,3,1,0,0,1,0,0,0,1,1,0,0,1,1,0,0,0,1,1,1,0,1,1,2,0,0,0,0,0,1,0,0,2,1,1,0,0,2,1,0,0,0,1,0,0,1,1,1,1,3,1,0,1,0,0,0,2,1,0,0,
0,1,0,0,1,0,0,0,0,0,0,2,3,0,0,2,0,1,0,3,0,0,1,2,0,1,0,0,1,1,0,1,0,0,0,1,1,0,0,0,1,0,0,1,0,0,0,1,2,0,0,0,2,2,0,1,1,0,1,1,3,1,0,2,0,1,0,2,2,0,0,1,1,0,0,1,1,0,0,0,0,0,1,1,2,0,0,0,0,1,0,1,0,1,1,1,1,1,0,0,
0,1,0,0,0,1,0,1,2,0,2,0,3,2,0,1,0,1,0,2,2,1,0,0,2,0,1,1,0,0,0,1,1,0,0,0,0,2,1,1,0,2,0,1,1,0,2,0,0,2,2,2,1,0,1,0,1,0,0,0,2,0,0,1,1,3,0,1,1,2,0,0,2,1,1,1,0,1,0,2,2,1,1,0,0,1,1,0,0,0,1,0,0,0,0,0,1,2,3,2,
0,0,1,0,0,1,0,0,0,1,0,1,2,0,0,0,1,0,0,0,0,0,0,1,1,0,1,3,2,1,1,0,0,0,0,0,1,1,2,0,0,1,0,0,0,1,0,1,0,1,0,0,1,0,0,1,1,0,0,1,0,1,0,3,2,2,1,2,0,2,0,0,0,0,0,0,1,1,0,0,0,1,1,0,0,1,0,2,1,1,1,0,0,3,0,0,2,1,1,2,
1,2,1,0,0,2,2,1,0,1,1,1,2,1,1,0,0,0,0,1,2,0,0,0,0,0,0,1,0,1,0,1,1,2,0,1,1,1,0,0,0,1,0,0,0,0,0,1,0,0,2,1,0,2,0,0,0,1,0,0,1,3,1,0,2,0,2,0,0,0,1,0,0,0,0,0,0,1,2,1,0,1,0,2,1,0,0,0,1,0,0,0,0,1,1,1,1,0,1,0,
1,0,1,1,0,0,1,2,0,0,0,1,0,1,0,1,1,0,2,4,2,1,0,0,0,1,0,1,0,1,1,1,0,0,1,1,1,1,0,2,0,0,0,1,3,0,0,0,0,2,0,1,2,0,0,0,0,0,1,1,1,2,1,1,2,1,0,3,0,0,1,0,0,2,1,0,1,0,2,1,0,0,1,1,1,1,0,2,2,1,0,2,1,2,0,2,0,0,1,1,
0,0,0,1,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,2,1,0,1,0,3,0,1,0,0,1,2,0,0,1,1,2,1,1,1,1,0,1,2,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0,1,1,2,0,0,0,3,2,1,3,0,0,0,1,1,0,0,0,1,1,1,2,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,2,
2,1,1,0,0,0,0,1,1,1,0,3,0,1,0,1,0,2,0,0,1,0,1,0,1,1,0,0,0,1,3,2,0,0,0,1,2,0,0,0,0,2,2,2,0,0,2,0,1,1,0,0,2,2,0,1,1,0,2,1,1,0,0,1,0,0,0,1,0,0,1,2,1,2,0,0,0,1,1,1,1,0,0,1,0,0,1,0,0,0,1,1,0,0,0,0,1,1,1,0,
1,1,0,1,0,0,1,1,2,1,2,0,0,2,0,2,1,1,0,0,0,0,0,0,0,1,0,3,1,1,0,1,0,1,1,0,0,1,0,1,0,1,3,1,0,0,1,0,1,0,2,2,0,1,0,0,1,1,1,0,0,1,0,0,0,1,0,2,0,0,0,1,2,2,1,1,0,1,1,0,0,1,1,0,0,0,2,1,2,1,0,1,0,0,0,1,2,1,0,3,
1,1,0,1,1,0,1,2,1,1,2,1,0,0,1,1,0,0,0,1,0,1,0,0,1,0,1,1,1,2,1,0,0,0,0,0,0,1,0,2,0,1,1,0,0,0,0,0,0,0,1,1,2,0,0,0,0,1,3,2,1,0,0,1,0,1,0,0,0,1,1,1,0,0,1,0,1,2,0,0,0,2,1,1,1,1,2,1,1,0,0,0,2,1,1,0,1,0,1,0,
1,1,1,1,1,1,1,0,1,0,0,1,0,0,0,0,0,3,1,2,1,1,0,2,1,0,1,0,2,1,0,0,0,0,0,0,1,1,0,0,0,1,1,1,0,0,0,1,0,3,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,3,0,0,1,2,0,0,0,1,0,3,2,0,0,0,2,0,0,0,0,0,1,3,0,3,1,0,2,0,1,1,0,1,
3,0,1,1,0,0,0,1,0,0,0,0,1,1,1,0,0,1,0,2,0,2,1,1,1,1,1,0,0,0,0,1,0,1,0,0,1,1,1,1,0,0,0,0,2,0,1,1,0,2,0,0,3,1,1,1,3,2,0,0,0,1,3,2,0,1,0,3,0,0,0,0,1,2,0,0,0,0,1,2,0,0,1,2,2,0,2,0,0,1,0,1,1,0,0,1,0,1,1,1,
0,0,0,0,0,0,1,0,0,0,1,1,1,1,0,0,1,1,1,0,1,3,0,1,1,0,2,1,0,2,1,1,0,1,0,1,0,1,1,2,0,2,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,0,3,0,0,0,0,0,2,0,0,2,0,1,3,0,0,1,1,2,1,0,2,1,1,1,1,0,1,0,3,0,0,0,1,0,0,0,
1,1,0,2,0,0,0,1,0,0,1,1,0,1,1,0,1,2,0,1,1,0,0,3,0,1,0,1,0,2,2,2,0,1,0,0,0,0,0,0,0,0,0,1,1,0,2,2,1,1,1,0,0,0,1,0,0,2,0,2,1,2,0,0,2,0,0,1,0,1,1,0,1,0,0,0,0,2,1,1,0,0,0,0,0,0,0,0,1,1,2,0,1,0,2,1,1,2,0,1,
2,1,0,0,0,1,0,1,0,0,0,0,0,1,0,2,1,0,1,1,0,1,0,0,0,1,3,0,0,0,0,0,1,2,0,0,1,1,1,1,0,0,0,1,1,0,0,1,3,1,1,1,0,0,1,1,0,1,0,0,1,1,0,0,0,1,2,1,2,2,0,0,0,2,1,0,0,1,0,1,0,1,0,0,1,2,0,2,0,0,0,1,1,0,0,0,1,1,0,0,
2,0,0,1,1,1,0,0,1,2,1,1,1,1,1,0,0,1,0,0,0,0,0,0,1,0,3,1,1,3,1,0,1,1,0,1,0,1,1,3,0,1,0,0,1,1,1,0,0,1,2,1,2,1,1,0,0,1,2,2,0,0,0,1,1,2,1,0,1,1,1,2,1,0,0,0,1,0,0,1,2,0,0,2,0,0,0,2,0,1,0,0,0,0,1,0,0,0,0,1,
0,2,0,0,0,1,0,3,1,0,1,0,2,0,1,2,0,0,0,0,2,1,1,1,2,0,0,0,0,2,0,1,0,0,0,1,1,1,0,1,0,1,1,0,0,0,0,0,3,2,0,0,0,1,0,2,0,1,1,1,1,0,0,0,1,0,0,1,2,0,2,2,0,1,1,0,1,0,1,1,1,0,0,2,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,1,
0};

int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        printf("%d\n",a[n]);
    }
}

Brian 2010-08-17 13:55 发表评论

POJ 1012 2244 Joseph 问题详解

Brian — Tue, 17 Aug 2010 05:42:00 GMT

约瑟夫环实在是太奇妙啦（我很高兴我的这篇原创文章被不少人转载了，虽然他们都没有引用出处... ...）！
1012 Joseph
Description

The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

Input

The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.

Output

The output file will consist of separate lines containing m corresponding to k in the input file.
2244 Eeny Meeny Moo

Description

Surely you have made the experience that when too many people use the Internet simultaneously, the net becomes very, very slow.
To put an end to this problem, the University of Ulm has developed a contingency scheme for times of peak load to cut off net access for some cities of the country in a systematic, totally fair manner. Germany's cities were enumerated randomly from 1 to n. Freiburg was number 1, Ulm was number 2, Karlsruhe was number 3, and so on in a purely random order.
Then a number m would be picked at random, and Internet access would first be cut off in city 1 (clearly the fairest starting point) and then in every mth city after that, wrapping around to 1 after n, and ignoring cities already cut off. For example, if n=17 and m=5, net access would be cut off to the cities in the order [1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7]. The problem is that it is clearly fairest to cut off Ulm last (after all, this is where the best programmers come from), so for a given n, the random number m needs to be carefully chosen so that city 2 is the last city selected.

Your job is to write a program that will read in a number of cities n and then determine the smallest integer m that will ensure that Ulm can surf the net while the rest of the country is cut off.

Input

The input will contain one or more lines, each line containing one integer n with 3 <= n < 150, representing the number of cities in the country.
Input is terminated by a value of zero (0) for n.

Output

For each line of the input, print one line containing the integer m fulfilling the requirement specified above.

1012打表做法 C :

#include
int a[14]={2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881,13482720};
int main()
{
int i;
while ( scanf("%d",&i), i != 0 )
printf("%d\n",a[i-1]);
return 0;
} // 这是从网上找的做法，号称打表法，这些数据依旧要通过建立循环链表或是别的模拟法来求出。但是单纯为了AC，这种做法真的是相当有效，讲白了就是有目的的穷举结果。

1012模拟法 C: 可惜呀可惜！这个总是超时！我不知道是什么原因。但是思路是正确的，可能有些地方我没有考虑到，看到这篇日志的人请指点。

#include

int main()
{
int i,m,k,cur,rest;

while(1)
{
  i=0; // the use ... sort of m in the question
  m=0;
  scanf("%d",&k);
  if (k == 0) break;
  while (1)
  {
   i++;
   rest=2*k; // good + bad guys
   cur=0;
   while (1)
   {
    cur=(cur+i-1)%rest; // find next from ZERO!
    if (cur >= k)
     rest--;
    else break;
   }
   if (rest == k)
   {
    m=i;
    break;
   }
  }
  printf("%d\n",m);
}
return 0;
}
对于 2244，建议看一个牛人的ACM博客: www.cppblog.com/AClayton/archive/2007/11/06/35964.html
我就是看这篇博文的，很牛的一个人 AClayton ，写的日期刚好是我生日，下面是其全部博文:
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在没有明白约瑟夫问题之前,只能用模拟来做.
      约瑟夫问题是这样的:
      假设n个人,编号为1到n,排成一个圈,顺时针从1开始数字m,数到m的人杀了,剩下的人继续游戏.活到最后的一个人是胜利者.一般来说需要编程求解最后一个人的编号.
      思路是这样的:
     假设当前剩下i个人(i<=n),显然这一轮m要挂(因为总是从1开始数).经过这一轮,剩下的人是:1 2 3 ... m- 1 m + 1 ... i, 我们将从m+1开始的数映射成1，则m+2对应2， n对应i - m， 1对应成i - m + 1  m - 1对应i - 1，那么现在的问题变成了已知i - 1个人进行循环报数m，求出去的人的序号。假设已经求出了i- 1个人循环报数下最后一个出去的人的序号X0，那么它在n个人中的序号X1=(X0+ m - 1) % n + 1, 最初的X0=1 ,反复迭代X0和X1可以求出.
     简单约瑟夫问题的解法:

#include <stdio.h >
main()
{
    int n, m,i, s=0;
    printf( "N  M  =  "); scanf("%d%d ",&n,&m);
    for(i=2;i<=n;i++)s=(s+m)%i;
    printf("The winner is %d\n ", s+1);
}

这倒题其实不是完全的约瑟夫问题,而是稍微变了形.呵呵,聪明的读者自己发现!这一点费了我很久的时间,还害我逃了课被点名...
这道题的我解法是这样的.

#include   <stdio.h >
int y(int n,int m)
{
    int s=1,i;
    for(i=2;i<=n-1;i++)
        s=(s+m-1)%i+1;
    return s+1;

}

main()
{
    int m,n;

    while(1)
    {
    scanf("%d",&n);
    if(n==0)break;
     for(m=1;;)
     {
         if(y(n,m)==2)break;
         m++;
     }


    printf("%d\n",m);

    }
}

读一个数处理一个数, Memory 68K,时间31MS,如果觉得效率不高. 优化的办法是打表~

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由此可见，将问题化归为数学问题，用初等高等或是数论来解决的能力是多么重要。

下面是我根据AClayton的思路简化后的代码，可直接AC: 注意，题目让你先让City 1 挂掉
2244 编译器 C :

#include<stdio.h>
void main()
{
int i,r,m,n;
while (scanf("%d",&n) && n)
{
    for (m=1; ; m++)
    {
     for (r=1,i=2; i<=n-1; i++)
      r=(r+m-1)%i + 1;
     if(r==1) break;
    }
    printf("%d\n",m);
}
} // 164K  16MS

Brian 2010-08-17 13:42 发表评论