hdu 5245 Joyful

Problem Description

Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like anM×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 theM×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 thet-th case, and then output the expected number of squares that will be painted. Round to integers.

 

Sample Input

23 3 14 4 2

 

Sample Output

Case #1: 4Case #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

题意：给一个N*M的网格，现在选K轮子矩形，每轮的矩形由随机选取的两个对顶点确定，求最终的被选中过的格子的数目。
分析：我们可以求出格子(i,j)被选中的概率pij,则答案为∑pij*i。

对于每个格子，k次选择相互独立，直接求选中的概率pij比较困难，但是k次未选中概率为pow(qij,k)比较好求，那么pij = 1 - pow(qij,k)。
但是对于其中一轮，格子(i,j)未被选中的qij不好求出，而选中概率比较好求（1-qij) = (1-((n-i)*(n-i)+(i-1)*(i-1)/(n*n)) * (1-((m-j)*(m-j)+(j-1)*(j-1))/(m*m));

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<set>#include<cmath>using namespace std;typedef long long LL;int main(){    int i,j,k,m,n,T;    int cas = 0;    scanf("%d",&T);    while(T--)    {        scanf("%d %d %d",&m,&n,&k);        double ans = 0;        for(i=1; i<=n; i++)            for(j=1; j<=m; j++)        {            double p = (1.0-(1.0*(n-i)*(n-i)+1.0*(i-1)*(i-1))/(1.0*n*n))                      *(1.0-(1.0*(m-j)*(m-j)+1.0*(j-1)*(j-1))/(1.0*m*m));            ans+=1.0-pow(1.0-p, k*1.0);        }        printf("Case #%d: %.0f\n",++cas,ans);    }    return 0;}