hdu 5245__Joyful

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 515    Accepted Submission(s): 224

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

 

題意：給出一個n*m的牆壁，進行k次染色，每次選取兩個點(x1,y1)，(x2,y2)作爲染色矩形的對角頂點。求k次染色後染色面積的期望值。

題解參考了這位大牛的博客。代碼如下。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    int T,n,m,k,cas=1;
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d%d",&n,&m,&k);
        double ans=0;
        for(double i=1;i<=n;i++) {
            for(double j=1;j<=m;j++) {
                double p=0;
                p+=(i-1)*(j-1)*(m-j+1)*(n-i+1);//1
                p+=(i-1)*(n-i+1)*m;//2
                p+=(i-1)*(m-j)*(n-i+1)*j;//3
                p+=(j-1)*(m-j+1)*n;//4
                p+=n*m;//5
                p+=(m-j)*n*j;//6
                p+=(n-i)*(j-1)*i*(m-j+1);//7
                p+=(n-i)*i*m;//8
                p+=(n-i)*(m-j)*i*j;//9
                p=p/n/m/n/m;
                ans+=1-pow(1-p,k);//k次染色操作被染色的概率
            }
        }
        printf("Case #%d: %d\n",cas++,int(ans+0.5));
    }
    return 0;
}