hdu5245——Joyful(概率论求期望)

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1198    Accepted Submission(s): 528

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.
题意：一个大矩形由m*n个小方块组成，一个人每次选两个小方块，他可以对以这两个方块为顶点的矩形面积内的小方块上色，一共可选k次，现在求被上色的小方块数目的期望
分析：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，用逆事件求小方格被上色的概率，遍历所有方格，不妨以当前方格为中心，那么只有四种情况，被选的两个小方块之间的区域同在中心的上下左右，不过这样会重复，四个角的矩形区域相当于被算了两次，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，做k次方，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double就行

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0xffffffff//0x3f3f3f3f
#define mod 100


const int M=1005;
ll n,m;
int cnt;
int sx,sy,sz;
int mp[1000][1000];
int pa[M*10],rankk[M];
int head[M*6],vis[M*100];
int dis[M*100];
int prime[M*1000];
bool isprime[M*1000];
int lowcost[M],closet[M];
char st1[5050],st2[5050];
//int len[M*6];
typedef pair<int ,int> ac;
int dp[55][55][55][55];
int has[1050000];
int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};
int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};

int main()
{
    int i,j,t;
    ll k;
    int cas=1;
    double ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&m,&n,&k);
        double tol=1.0*(n*m)*(m*n);
        ans=0.0;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                double tmp=0;
                tmp+=1.0*(i-1)*n*(i-1)*n+1.0*(j-1)*m*(j-1)*m+1.0*(m-i)*n*(m-i)*n+1.0*(n-j)*m*(n-j)*m;
                tmp-=1.0*(i-1)*(i-1)*(j-1)*(j-1)+1.0*(i-1)*(i-1)*(n-j)*(n-j)+1.0*(j-1)*(j-1)*(m-i)*(m-i)+1.0*(m-i)*(m-i)*(n-j)*(n-j);
                double p=1.0*tmp/tol;
                p=pow(p,k);
                ans+=(1.0-p);
            }
        }
        printf("Case #%d: %d\n",cas++,(int)(ans+0.5));
    }
    return 0;
}