hdu 5245 Joyful（期望）

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1243    Accepted Submission(s): 546

Problem Description

Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an M×N matrix. The wall has M×N squares in all. In the whole problem we denotes (x,y) to be the square at the x-th row, y-th column. Once Sakura has determined two squares (x1,y1) and (x2,y2), she can use the magical tool to paint all the squares in the sub-matrix which has the given two squares as corners.

However, Sakura is a very naughty girl, so she just randomly uses the tool for K times. More specifically, each time for Sakura to use that tool, she just randomly picks two squares from all the M×N squares, with equal probability. Now, kAc wants to know the expected number of squares that will be painted eventually.

 

Input

The first line contains an integer T(T≤100), denoting the number of test cases.

For each test case, there is only one line, with three integers M,N and K.
It is guaranteed that 1≤M,N≤500, 1≤K≤20.

 

Output

For each test case, output ''Case #t:'' to represent the t-th case, and then output the expected number of squares that will be painted. Round to integers.

 

Sample Input

2 3 3 1 4 4 2

 

Sample Output

Case #1: 4 Case #2: 8

Hint

The precise answer in the first test case is about 3.56790123.

 

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

 

Recommend

 

 

直接求好像不大好求，可以反着求

代码较挫，

#include <iostream>
#include <stdio.h>
using namespace std;

using namespace std;

#define LL long long

long long C(LL  a,LL b)
{
    return a * b * a * b;
}
int main()
{
    long long T,m,n,k;
    int cas = 0;
    scanf("%lld",&T);
    while(T--)
    {
        LL sum;
        double ans=0, p;
        scanf("%lld%lld%lld",&m,&n,&k);
        for(LL i=0; i<m; i++)
        {
            for(LL j=0; j<n; j++)
            {
               sum = C(i,n)+C((m-i-1),n)+C(j,m)+C((n-j-1),m);
               sum = sum - C(i,j) - C(i,(n-j-1)) - C((m-i-1),j ) - C((m-i-1),(n-j-1)) ;
               p = sum * 1.0 / C(m,n);

double tmp=1;
               for (LL jj = 1; jj <= k; ++jj)
                    tmp *= p;
               ans=ans+1-tmp;
            }
        }
        printf("Case #%d: %.0f
",++cas,ans);
    }
    return 0;
}