C++博客-Vontroy-随笔分类-HDU

HDU 2734 Quicksum 简单字符串处理

Vontroy — Sun, 03 Oct 2010 02:03:00 GMT

Quicksum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 492 Accepted Submission(s): 408

Problem Description

A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for detecting transmission errors, validating document contents, and in many other situations where it is necessary to detect undesirable changes in data.

For this problem, you will implement a checksum algorithm called Quicksum. A Quicksum packet allows only uppercase letters and spaces. It always begins and ends with an uppercase letter. Otherwise, spaces and letters can occur in any combination, including consecutive spaces.

A Quicksum is the sum of the products of each character's position in the packet times the character's value. A space has a value of zero, while letters have a value equal to their position in the alphabet. So, A=1, B=2, etc., through Z=26. Here are example Quicksum calculations for the packets "ACM" and "MID CENTRAL":

ACM: 1*1 + 2*3 + 3*13 = 46MID CENTRAL: 1*13 + 2*9 + 3*4 + 4*0 + 5*3 + 6*5 + 7*14 + 8*20 + 9*18 + 10*1 + 11*12 = 650

Input

The input consists of one or more packets followed by a line containing only # that signals the end of the input. Each packet is on a line by itself, does not begin or end with a space, and contains from 1 to 255 characters.

Output

For each packet, output its Quicksum on a separate line in the output.

Sample Input

ACM
MID CENTRAL
REGIONAL PROGRAMMING CONTEST
ACN
A C M
ABC
BBC
#

Sample Output

#include <iostream>
#include <cstdio>
#include <cstring>

int main()
{
    char str[260];
    int ans = 0;
    while( gets(str) && str[0] != '#' )
    {
        ans = 0;
        int len = strlen(str);
        for( int i = 0; i < len; i++ )
        {
            if( str[i] == ' ' )
                continue;
            else
                ans += ( str[i] - 64 ) * ( i + 1 );
        }
        std::cout << ans << std::endl;
    }
    return 0;
}

Vontroy 2010-10-03 10:03 发表评论

HDU 3661 Assignments-2010 Harbin Regional

Vontroy — Sun, 03 Oct 2010 00:36:00 GMT

Assignments

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 252 Accepted Submission(s): 125

Problem Description

In a factory, there are N workers to finish two types of tasks (A and B). Each type has N tasks. Each task of type A needs xi time to finish, and each task of type B needs yj time to finish, now, you, as the boss of the factory, need to make an assignment, which makes sure that every worker could get two tasks, one in type A and one in type B, and, what's more, every worker should have task to work with and every task has to be assigned. However, you need to pay extra money to workers who work over the standard working hours, according to the company's rule. The calculation method is described as follow: if someone’ working hour t is more than the standard working hour T, you should pay t-T to him. As a thrifty boss, you want know the minimum total of overtime pay.

Input

There are multiple test cases, in each test case there are 3 lines. First line there are two positive Integers, N (N<=1000) and T (T<=1000), indicating N workers, N task-A and N task-B, standard working hour T. Each of the next two lines has N positive Integers; the first line indicates the needed time for task A1, A2…An (Ai<=1000), and the second line is for B1, B2…Bn (Bi<=1000).

Output

For each test case output the minimum Overtime wages by an integer in one line.

Sample Input

2 5
4 2
3 5

Sample Output

//简单贪心
#include <iostream>
#include <cstdio>
#include <algorithm>

const int maxn = 1000 + 5;

using namespace std;

bool cmp( int a, int b )
{
    return a > b;
}

int main()
{
    int a[maxn], b[maxn];
    int n, t, ans = 0;
    while( cin >> n >> t )
    {
        ans = 0;
        for( int i = 0; i < n; i++ )
            cin >> a[i];
        for( int i = 0; i < n; i++ )
            cin >> b[i];

        sort( a, a + n );
        sort( b, b + n, cmp );

        for( int i = 0; i < n; i++ )
            if( a[i] + b[i] > t )
                ans += a[i] + b[i] - t;
        cout << ans << endl;
    }
    return 0;
}

Vontroy 2010-10-03 08:36 发表评论

HDU 1097 A hard puzzle

Vontroy — Sat, 02 Oct 2010 16:39:00 GMT

//找规律，每4个数循环一次
#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    long long a, b;
    int ans[5];
    while(cin >> a >> b)
    {
        if( b == 0 )
        {
            cout << "1" << endl;
            continue;
        }
        else if( a == 0 )
        {
            cout << "0" << endl;
            continue;
        }
        ans[1] = a % 10;
        ans[2] = (ans[1] * a ) % 10;
        ans[3] = (ans[2] * a ) % 10;
        ans[4] = (ans[3] * a ) % 10;

        int tmp = b % 4;
        if( tmp == 0 ) tmp = 4;
        cout << ans[tmp] << endl;
    }
    return 0;
}

Vontroy 2010-10-03 00:39 发表评论

HDU 1452 Happy 2004(因子和)

Vontroy — Sat, 02 Oct 2010 14:08:00 GMT

/*************************************************************
计算 2004^X的因子和 s(2004^X)   mod M, M=29

  s(2004^X)%29
  因子和 s是积性函数,即：gcd(a,b)=1==> s(a*b)= s(a)*s(b)

  2004^X=4^X * 3^X *167^X
  s(2004^X)=  s(2^(2X))* s(3^X) * s(167^X)

  如果 p是素数 ==> s(p^X)=1+p+p^2++p^X = (p^(X+1)-1) /(p-1)

  s(2004^X)=（2^(2X+1)-1）* (3^(X+1)-1)/2  *(167^(X+1)-1)/166

  167%29=22

  s(2004^X)=（2^(2X+1)-1）* (3^(X+1)-1)/2  *(22^(X+1)-1)/21

  (a*b)/c %M= a%M* b%M  * inv(c)

  c*inv(c)=1 %M    模为1的所有数  inv（c）为最小可以被c整除的

    inv(2)=15,  inv(21)=18    2*15=1 mod 29, 18*21=1 mod 29

  s(2004^X)=（2^(2X+1)-1）* (3^(X+1)-1)/2  *(22^(X+1)-1)/21
           =（2^(2X+1)-1）* (3^(X+1)-1)*15 *(22^(X+1)-1)*18
***************************************************************/

//hdu 1452
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int _pow( int a, int n )
{
    int b = 1;
    while( n > 1 )
        if( n % 2 == 0 )
        {
            a = ( a * a ) % 29;
            n /= 2;
        }
        else
        {
            b = b * a % 29;
            n--;
        }
        return a * b % 29;
}
int main()
{
    int X;
    int a, b, c;
    while( cin >> X, X )
    {
        a = _pow( 2, 2 * X + 1 );
        b = _pow( 3, ( X + 1 ) );
        c = _pow( 22, ( X + 1 ) );

        cout << ( a - 1 ) * ( b - 1 ) * 15 * ( c - 1 ) * 18 % 29 << endl;
    }
    return 0;
}

Vontroy 2010-10-02 22:08 发表评论

HDU 1056 HangOver

Vontroy — Sat, 02 Oct 2010 08:08:00 GMT

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    double len;
    double tmp;
    int ans;

    while( cin >> len, len )
    {
        tmp = 0;
        ans = 1;
        while( ans++ )
        {
            tmp += 1 / ( ans * 1.0 );
            if( tmp >= len )
            {
                cout << ans - 1 << " card(s)" << endl;
                break;
            }
        }
    }
    return 0;
}

Vontroy 2010-10-02 16:08 发表评论

HDU 1788 Chinese remainder theorem again

Vontroy — Sat, 02 Oct 2010 06:58:00 GMT

/**********************************
N % MI = MI - a
因为 a < MI
原式等价于 (N + a) % MI = 0
所以此题为求 M0 到 MI 的最小公倍数

(注意精度问题,用__int64)

***********************************/

#include <iostream>
#include <cstdio>

using namespace std;

__int64 gcd( __int64 a, __int64 b )
{
    if( b == 0 ) return a;
    return gcd( b, a % b );
}

__int64 lcm( __int64 a, __int64 b )
{
    return a * b / gcd( a, b );
}

int main()
{
    int n, k;
    int tmp;
    __int64 ans;
    while( cin >> n >> k, n || k )
    {
        ans = 1;
        for( int i = 0; i < n; i++ )
        {
            cin >> tmp;
            ans = lcm( ans, tmp );
        }
        cout << ans - k << endl;
    }
    return 0;
}

Vontroy 2010-10-02 14:58 发表评论

HDU 1018 Big Number

Vontroy — Sat, 02 Oct 2010 06:22:00 GMT

斯特灵公式是一条用来取n阶乘近似值的数学公式。一般来说，当n很大的时候，n阶乘的计算量十分大，所以斯特灵公式十分好用，而且，即使在

n很小的时候，斯特灵公式的取值已经十分准确。

公式为：

这就是说，对于足够大的整数n，这两个数互为近似值。更加精确地：

或者：

Big Number

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8759 Accepted Submission(s): 3879

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input

2
10
20

Sample Output

7
19

//log10(n!)=(0.5*log(2*PI*n)+n*log(n)-n)/log(10)

#include <iostream>
#include <cstdio>
#include <cmath>

const double PI = 3.1415926;

int main()
{
    int n;
    int tmp;
    while( ~scanf("%d", &n ) )
    {
        for( int i = 0; i < n; i++ )
        {
            scanf("%d", &tmp);
            double cnt = 1;
            cnt += (0.5 * log( 2 * PI * tmp ) + tmp * log( tmp ) - tmp ) / log(10);
            printf("%d\n", (int)(cnt));
        }
    }
    return 0;
}

Vontroy 2010-10-02 14:22 发表评论

HDU 1316 How Many Fibs?

Vontroy — Sat, 02 Oct 2010 06:00:00 GMT

比较水的大数。。。
代码：

import java.util.*;
import java.math.*;

public class Main{
    public static void main( String args[] )
    {
        Scanner in = new Scanner( System.in );
        int cnt = 0;
        BigInteger fib1, fib2;
        BigInteger fla1, fla2;
        while( in.hasNext() )
        {
            fib1 = BigInteger.valueOf(1);
            fib2 = BigInteger.valueOf(2);

            fla1 = in.nextBigInteger();
            fla2 = in.nextBigInteger();
            if( fla1.equals(BigInteger.valueOf(0)) && fla2.equals(BigInteger.valueOf(0))) break;

            cnt = 0;

            while( true )
            {
                if( fib1.compareTo(fla1) >= 0 && fib1.compareTo(fla2) <= 0 ) cnt++;
                if( fib2.compareTo(fla1) >= 0 && fib2.compareTo(fla2) <= 0 ) cnt++;
                if( fib1.compareTo(fla2) > 0 || fib2.compareTo(fla2) > 0 ) break;
                fib1 = fib1.add(fib2);
                fib2 = fib2.add(fib1);
            }
            System.out.println( cnt );
        }
    }
}

Vontroy 2010-10-02 14:00 发表评论

HDU 1198 Farm Irrigation

Vontroy — Wed, 28 Jul 2010 23:04:00 GMT

#include<iostream>
using namespace std;

int bin[2500];

int find(int x)
{return x==bin[x]?x:find(bin[x]);}

void merge(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x!=y) bin[x]=y;
}

int main()
{
    int a[11] = {9, 3, 12, 6, 5, 10, 11, 13, 14, 7, 15}, m, n, i, j, map[50][50], total;
    char c;
    while(scanf("%d %d", &m, &n), m != -1 || n != -1){
        getchar();
        for(i = 0; i < m; ++i){
            for(j = 0; j < n; ++j){
                scanf("%c", &c);
                map[i][j] = c - 65;
            }
            getchar();
        }
        for(i = 0; i < m * n; ++i) bin[i] = i;
        for(i = 0; i < m; ++i)
            for(j = 0; j < n; ++j){
                if(i + 1 < m && a[map[i][j]] & 0x04 && a[map[i + 1][j]] & 0x01)
                    merge(i * n + j, i * n + n + j);
                if(j + 1 < n && a[map[i][j]] & 0x02 && a[map[i][j + 1]] & 0x08)
                    merge(i * n + j, i * n + j + 1);
            }
        for(total = 0, i = 0; i < m * n; ++i)
            if(i == bin[i]) ++total;
        printf("%d\n", total);
    }
    return 0;
}

Vontroy 2010-07-29 07:04 发表评论

HDU 1010 Tempter of the Bone

Vontroy — Wed, 28 Jul 2010 22:58:00 GMT

#include <iostream>
#include <cstdio>
#include <cmath>

const int maxn = 10;

using namespace std;

bool escape;

char map[maxn][maxn];

int dir[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};
int n, m, t;
int si, sj, di, dj;

void dfs( int x, int y, int cnt )
{
    if( escape ) return;

    if( x == di && y == dj && cnt == t )
    {
        escape = 1;
        return;
    }

    int temp = t - cnt - ( fabs(di - x) + fabs(dj - y ) );

    if( temp < 0 || temp % 2 != 0 )     return;//奇偶剪枝

    if( x <= 0 || y <= 0 || x > n || y > m )    return;

    for( int i = 0; i < 4; i++ )
    {
        if( map[x+dir[i][0]][y+dir[i][1]] != 'X')
        {
            map[x+dir[i][0]][y+dir[i][1]] = 'X';
            dfs(x+dir[i][0], y+dir[i][1], cnt+1);
            map[x+dir[i][0]][y+dir[i][1]] = '.';
        }
    }

}

int main()
{
    while( cin >> n >> m >> t, m || n || t )
    {
        int wall = 0;
        escape = 0;
        for ( int i = 1; i <= n; i++ )
            for ( int j = 1; j <= m; j++ )
            {
                cin >> map[i][j];

                if( map[i][j] == 'D')
                {
                    di = i;
                    dj = j;
                }

                else if ( map[i][j] == 'S' )
                {
                    si = i;
                    sj = j;
                }

                else if ( map[i][j] == 'X' )
                    wall ++ ;
            }

        if( m * n - wall <= t )
        {
            printf("NO\n");
            continue;
        }
        else
        {
            map[si][sj] = 'X';
            dfs( si, sj, 0 );
            printf("%s\n", escape ? "YES" : "NO" );
        }
    }
    return 0;
}

Vontroy 2010-07-29 06:58 发表评论