HDU 5245 Joyful（概率 期望）

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1295    Accepted Submission(s): 567

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

题意：

题意大致是：进行K次染色，每次染色会随机选取一个以(x1,y1),(x2,y2)为一组对角的子矩阵进行染色，求K次染色后染色面积的期望值(四舍五入)。

开始还在想通过两个端点来想，但是由于可以k次染色，所以分割来看简单：

摘抄网上（和我之前想的思路是一样的，）：

每个格子可以被重复选，因此可以把每一个小方块选不选当做一个独立事件，所以我们算出每个小方块对总期望的贡献值就行了，直接算不好算，考虑算每个小方块一次不被选的概率p，这个算起来就方便很多，不妨以当前点为中心，那么只有四种情况，被选的两个小方块同在中心的上下左右，不过这样会重复，四个角相当于被算了两次，再把它们减掉一次，这样求出来的是当前点一次不会被上色的情况数，除以总情况数(m * n * m * n)就是当前方块一次不会被上色的概率，再乘k次，就是k次这个方块都不被上色的概率，用1减则为选k次这个方块被上色的概率，只需要把每个方块选k次后被上色的概率累加即可，情况数中间可能会爆int，用long long

只不过在我看是看到这道题目的时候，犯了一个致命的错误，就是把分割的那四块计算时本来是可以重复选的，也就是起始和另外一个端点是可以同一个端点的，但是我竟然想成组合问题了，然后C(n,i)然后就错了。。。不过还好想过来了。

我地AC代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{

    long long  T,m,n,k;
    scanf("%d",&T);
    int Case=1;
    while(T--)
    {
        double ans=0;
        long long sum;
        scanf("%lld%lld%lld",&m,&n,&k);
        for(long long  i=0; i<m; i++)
        {
            for(long long j=0; j<n; j++){
            sum=0;
               sum=sum+i*n*i*n;
               sum=sum+(m-i-1)*n*(m-i-1)*n;
               sum=sum+j*m*j*m;
               sum=sum+(n-j-1)*m*(n-j-1)*m;
               sum=sum-i*j*i*j;
               sum=sum-i*(n-j-1) * i*(n-j-1);
               sum=sum-(m-i-1)*j * (m-i-1)*j;
               sum=sum-(m-i-1)*(n-j-1) * (m-i-1)*(n-j-1);

               ans=ans+1- pow(sum*1.0 / ((m*n)*(m*n)) ,k);
            }
//            cout << sum << endl;
        }
        printf("Case #%d: %.0f
",Case++,ans);
    }
    return 0;
}