HDU 5245 Joyful（概率求期望）——2015年上海邀请赛

版权声明：编写不易，转载请注明出处，谢谢。 https://blog.csdn.net/qingshui23/article/details/78356547

传送门

Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an $M \times N$ matrix. The wall has $M \times N$ 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 $(x_1, y_1)$ and $(x_2, y_2)$, 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 $M \times N$ 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 $T$( $T \le 100$), 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 $1 \le M, N \le 500$, $1 \le K \le 20$.

 

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.
    
    
   
   

题目大意

给定一个 $n*m$ 的方格，每次选取两个格子$a(x_1,y_1)$，$b(x_2,y_2)$ 以这两个格子作为矩形的对角点，得到一个矩形，然后将其涂上颜色，求涂 $k$ 次之后得到的有颜色的面积的期望。

解题思路

就是对于每一个格子，求其 $k$ 次之后被染色的概率，然后将每一个格子的概率加起来就是期望。
然而对于每一个格子来说，求 $k$ 次之后被染色的概率不好求，我们可以求 $k$ 次都没有被染色的概率 $p$，然后 $1-p$ 就是我们所求的概率。怎么求不被染色的概率呢，其实就是以当前所求点 $(i,j)$ 为中心，分为四个部分。每个部分的面积分别为：
左部分：$(j-1)*n$
右部分：$(m-j)*n$
上部分：$(i-1)*m$
下部分：$(n-i)*m$
再减去重复的部分：
左上：$(i-1)*(j-1)$
右上：$(i-1)*(m-j)$
左下：$(n-i)*(j-1)$
左下：$(n-i)*(m-j)$
所以我们得到总的没有被涂色的面积为 $tmp$，涂色的面积就是 $1-{\frac {tmp}{n*m*n*m}}^k$

代码

#include <iostream>
#include <string.h>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
const double PI = acos(-1);
const double eps = 1e-8;
const LL MOD = 1e9+7;
int main(){
    //freopen("C:/Users/yaonie/Desktop/in.txt", "r", stdin);
    //freopen("C:/Users/yaonie/Desktop/out.txt", "w", stdout);
    int T; scanf("%d",&T);
    for(int cas=1; cas<=T; cas++){
        LL n, m, k; scanf("%lld%lld%lld",&n,&m,&k);
        double ans = 0;
        LL t = n*n*m*m;
        for(LL i=1; i<=n; i++){
            for(LL j=1; j<=m; j++){
                LL tmp = (j-1)*(j-1)*n*n-(i-1)*(i-1)*(j-1)*(j-1);
                tmp += ((m-j)*(m-j)*n*n-(i-1)*(i-1)*(m-j)*(m-j));
                tmp += ((i-1)*(i-1)*m*m-(n-i)*(n-i)*(j-1)*(j-1));
                tmp += ((n-i)*(n-i)*m*m-(n-i)*(n-i)*(m-j)*(m-j));
                double sum = 1;
                double a = 1.0*tmp/t;
                LL b = k;
                while(b){
                    if(b&1) sum=sum*a;
                    b>>=1;
                    a=a*a;
                }
                ans += 1-sum;
            }
        }
        printf("Case #%d: %.0f\n",cas, ans);
    }
    return 0;
}