C++博客-ACM___________________________-随笔分类-ACM ( 数学题/几何 )

HDU 1098 HDOJ 1098 Ignatius's puzzle ACM 1098 IN HDU

MiYu — Thu, 28 Oct 2010 13:17:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1098

题目分析:

纯粹的数学题, 数学归纳法的应用 , 最后归纳得出原题等价与 18 + k*a 是否能被65整除.

代码如下 :

代码/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_1098
Doc Name  : Ignatius's puzzle
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num)
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
int main ()
{
    int N;
    while ( scan_d ( N ) ) {
        if ( N % 65 == 0 )
            puts ( "no" );
        else {
            int i;
            for ( i = 0; i < 65; ++ i ) {
                if ( ( i * N + 18 ) % 65 == 0 ) {
                    printf ( "%d\n", i );
                    break;
                }
            }
            if ( i == 65 ) puts ( "no" );
        }
    }
    return 0;
}

MiYu 2010-10-28 21:17 发表评论

HDOJ 1999 HDU 1999 不可摸数 ACM 1999 IN HDU

MiYu — Sun, 15 Aug 2010 10:00:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1999
题目描述:

不可摸数

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2613 Accepted Submission(s): 694

Problem Description
s(n)是正整数n的真因子之和，即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何
数m，s(m)都不等于n,则称n为不可摸数.

Input
包含多组数据，首先输入T,表示有T组数据.每组数据1行给出n(2<=n<=1000)是整数。

Output
如果n是不可摸数，输出yes，否则输出no

Sample Input
3
2
5
8

Sample Output
yes
yes
no

题目分析:

标准的筛选法。求出每个数的因子和，

然后看因子和是否在1000以内，是的话就证明等于因子和的这个数是不可摸数。

代码如下: ( 奋斗哥代码 0rz................... )

#include <iostream>
#include <string.h>
#include <cmath>
using namespace std;

int sum[1000001], sign[1001];
int main()
{
    int nCases, num;
    scanf("%d", &nCases);
    for(int i = 1; i <= 500000; ++i)
        for(int j = 2*i; j <= 1000000; j += i)
            sum[j] += i;
    for(int i = 1; i <= 1000000; ++i)
        if(sum[i] < 1000)
            sign[sum[i]] = 1;

    while(nCases--)
    {
        scanf("%d", &num);
        if(sign[num])
            printf("no\n");
        else
            printf("yes\n");
    }
    return 0;
}

MiYu 2010-08-15 18:00 发表评论

HDOJ 2036 HDU 2036 改革春风吹满地 ACM 2036 IN HDU

MiYu — Thu, 12 Aug 2010 03:19:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2036
题目描述:

改革春风吹满地

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5623 Accepted Submission(s): 2763

Problem Description
“ 改革春风吹满地,
不会AC没关系;
实在不行回老家，
还有一亩三分地。
谢谢!（乐队奏乐）”

话说部分学生心态极好，每天就知道游戏，这次考试如此简单的题目，也是云里雾里，而且，还竟然来这么几句打油诗。
好呀，老师的责任就是帮你解决问题，既然想种田，那就分你一块。
这块田位于浙江省温州市苍南县灵溪镇林家铺子村，多边形形状的一块地，原本是linle 的，现在就准备送给你了。不过，任何事情都没有那么简单，你必须首先告诉我这块地到底有多少面积，如果回答正确才能真正得到这块地。
发愁了吧？就是要让你知道，种地也是需要AC知识的！以后还是好好练吧

Input
输入数据包含多个测试实例，每个测试实例占一行，每行的开始是一个整数n(3<=n<=100)，它表示多边形的边数（当然也是顶点数），然后是按照逆时针顺序给出的n个顶点的坐标（x1, y1, x2, y2 xn, yn）,为了简化问题，这里的所有坐标都用整数表示。
输入数据中所有的整数都在32位整数范围内，n=0表示数据的结束，不做处理。

Output
对于每个测试实例，请输出对应的多边形面积，结果精确到小数点后一位小数。
每个实例的输出占一行。

Sample Input
3 0 0 1 0 0 1
4 1 0 0 1 -1 0 0 -1
0

Sample Output
0.5
2.0

题目分析:

模板题. 直接用多边形面积公式 : S = 1/2 * abs( ∑(x_iy_i+1 – x_i+1y_i) )

代码如下:

#include <iostream>
#include <cmath>
using namespace std;
typedef struct point{
int x, y;
}point;
point polygon[101];

double areaGmty ( int num )
{
       double area = 0;
       for ( int i = 0; i < num; ++ i )
       {
              area += ( polygon[i].x * polygon[(i+1)%num].y ) - (polygon[(i+1)%num].x * polygon[i].y);
       }
       return area;
}
int main ()
{
    int N;
    while ( scanf ( "%d",&N ), N )
    {
          for ( int i = 0; i != N; ++ i )
          {
                scanf ( "%d%d", &polygon[i].x,&polygon[i].y );
          }
          double area = areaGmty ( N ) / 2.0;
          printf ( "%.1lf\n",area );
    }
    return 0;
}

MiYu 2010-08-12 11:19 发表评论

HDOJ 1969 HDU 1969 Pie ACM 1969 IN HDU

MiYu — Wed, 11 Aug 2010 06:37:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1969
题目描述:

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 65

Problem Description
My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input
One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input
3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output
25.1327
3.1416
50.2655

题目分析:

2分求解.

题目大意是要办生日Party，有n个馅饼，有f个朋友，接下来是n个馅饼的半径。然后是分馅饼了，
注意咯自己也要，大家都要一样大，形状没什么要求，但都要是一整块的那种，也就是说不能从两个饼中
各割一小块来凑一块，像面积为10的和6的两块饼（饼的厚度是1，所以面积和体积相等），
如果每人分到面积为5，则10分两块，6切成5，够分3个人，如果每人6，则只能分两个了！
题目要求我们分到的饼尽可能的大！

只要注意精度问题就可以了,一般WA 都是精度问题.

代码如下:

#include <iostream>
#include <cmath>
using namespace std;
double a[10005];
int N,f;
double pi = acos ( -1.0 );
bool test ( double x )
{
    int count = 0;
    for ( int i = 1; i <= N; ++ i )
    {
        count += int ( a[i] / x );
    }
    if ( count >= f + 1 )
    {
         return true;
    }
    else
    {
         return false;
    }
}
int main()
{
    int T;
    scanf ( "%d", &T );
    while ( T -- )
    {
           double sum = 0;
           double rad;
           scanf("%d%d",&N,&f);
           for ( int i = 1; i <= N; ++ i )
           {
               scanf ( "%lf", &rad );
               a[i] = rad * rad * pi;
               sum += a[i];
           }
           double max = sum / ( f + 1 );
           double l = 0.0;
           double r = max;
           double mid = 0.0;
           while ( r - l > 1e-6 )
           {
                  mid = ( l + r ) / 2;
                  if ( test ( mid ) )
                  {
                       l = mid;
                  }
                  else
                  {
                       r = mid;
                  }
           }
           printf("%.4lf\n",mid);
    }
    return 0;
}

MiYu 2010-08-11 14:37 发表评论

HDOJ 2899 HDU 2899 Strange fuction ACM 2899 IN HDU

MiYu — Wed, 11 Aug 2010 05:51:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2899
题目描述:

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95    Accepted Submission(s): 75

Problem Description
Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input
2
100
200

Sample Output
-74.4291
-178.8534

题目分析:

纯数学题, 需要微积分的知识分析题目...........我承认....我数学没怎么学好..........
分析出来使用2分搜索就可以了..........

代码如下:

#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
       return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
    int T;
    scanf ( "%d",&T );
    while ( T -- )
    {
          scanf ( "%lf",&y );
          if ( cal(0) > y || cal(100) < y )
          {
               printf ( "No solution!\n" );
               continue;
          }
          double l = 0.0, r = 100.0,res = 0.0;
          while ( r - l > 1e-6 )
          {
                double mid = ( l + r ) / 2.0;
                res = cal ( mid );
                if ( res > y )
                     r = mid - 1e-6;
                else
                     l = mid + 1e-6;
          }
          printf ( "%.4lf\n",( l + r ) / 2.0 );
    }
    return 0;
}

MiYu 2010-08-11 13:51 发表评论

HDOJ HDU 2073 无限的路 ACM 2073 IN HDU

MiYu — Sat, 07 Aug 2010 08:25:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2073
题目描述:

Problem Description
甜甜从小就喜欢画图画，最近他买了一支智能画笔，由于刚刚接触，所以甜甜只会用它来画直线，于是他就在平面直角坐标系中画出如下的图形：

甜甜的好朋友蜜蜜发现上面的图还是有点规则的，于是他问甜甜：在你画的图中，我给你两个点，请你算一算连接两点的折线长度（即沿折线走的路线长度）吧。

Input
第一个数是正整数N（≤100）。代表数据的组数。
每组数据由四个非负整数组成x1，y1，x2，y2；所有的数都不会大于100。

Output
对于每组数据，输出两点(x1,y1),(x2,y2)之间的折线距离。注意输出结果精确到小数点后3位。

Sample Input
5
0 0 0 1
0 0 1 0
2 3 3 1
99 99 9 9
5 5 5 5

Sample Output
1.000
2.414
10.646
54985.047
0.000

题目分析:
简单的数学题, 只需要把2点之间的所有线段加进去就可以了 .

代码如下:

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int N, x, x0, x1, y, y0, y1;
    cin >> N;
    while ( N-- )
    {
            cin >> x0 >> y0 >> x1 >> y1;
            if ( x0 + y0 > x1 + y1 )
            {
                 x0 ^= x1 ^= x0 ^= x1;
                 y0 ^= y1 ^= y0 ^= y1;
            }
            x = x0 + y0;
            y = x1 + y1;
            double line = sqrt ( pow ( x0 - x1 * 1.0, 2 ) + pow ( y0 - y1 * 1.0, 2 ) ); // 2点在同一直线上
            double len = sqrt ( 2.0 ) * ( x + x1 - x0 + y * ( y - 1.0 ) / 2.0 - x * ( x + 1.0 ) / 2.0 );  //所有斜率为1的直线的长度和
            for ( int i = x; i < y; ++ i )
            {
                  len += sqrt ( 2.0 * i * i + 2.0 * i + 1 );  //x 与 y 之间的全部斜线(斜率不为1)相加
            }
            printf( "%.3f\n", x == y ? line : len );
    }
    return 0;
}

MiYu 2010-08-07 16:25 发表评论