hdu 5245 Joyful(期望的计算，好题)

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

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)。

对于这道题的话，首先要考虑的是进行一次选择时的期望。求期望的方法为单独考虑每一格所能获得的期望，然后将所有格的期望相加即为答案。

对于每一个所能获得的期望，即要计算所有包含这一格的个数ans，除于总的选择方案tot

此时我们的问题转向了如何计算A[x.y]上

由题目描述，一次染色中可能的操作有n^2*m^2种

计算A[x,y]时，我们可以把整个矩阵做如下拆分

当前计算的方块为[x,y]，即图中编号为5的部分

将其他部分拆分成图上8个区域，则可得到以下关系

对于一种染色方案能够覆盖方块[x,y]时
①[x1,y1]取在区域1内时，[x2,y2]可以在5、6、8、9四个区域内任取;
②[x1,y1]取在区域2内时，[x2,y2]可以在4、5、6、7、8、9六个区域内任取;
③[x1,y1]取在区域3内时，[x2,y2]可以在4、5、7、8四个区域内任取;
④[x1,y1]取在区域4内时，[x2,y2]可以在2、3、5、6、8、9六个区域内任取;
⑤[x1,y1]取在区域5内时，[x2,y2]可以在所有区域内任取;
⑥[x1,y1]取在区域6内时，[x2,y2]可以在1、2、4、5、7、8六个区域内任取;
⑦[x1,y1]取在区域7内时，[x2,y2]可以在2、3、5、6四个区域内任取;
⑧[x1,y1]取在区域8内时，[x2,y2]可以在1、2、3、4、5、6六个区域内任取;
⑨[x1,y1]取在区域1内时，[x2,y2]可以在1、2、4、5四个区域内任取;

计算出这个格子的概率p后，总的答案加上 1-pow（1-p，k），得到最后的答案

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<stdlib.h>
 6 #include<queue>
 7 using namespace std;
 8 #define ll long long
 9 int m,n,k;
10 int main()
11 {
12     int t;
13     int ac=0;
14     scanf("%d",&t);
15     while(t--){
16         scanf("%d%d%d",&n,&m,&k);
17         double ans=0;
18         for(int i=1;i<=n;i++){
19             for(int j=1;j<=m;j++){
20                 double tmp=0;
21                 tmp=tmp+(double)(i-1)*(j-1)*(n-i+1)*(m-j+1);//1
22                 tmp=tmp+(double)(i-1)*(n-i+1)*m;//2
23                 tmp=tmp+(double)(i-1)*(m-j)*(n-i+1)*j;//3
24                 tmp=tmp+(double)(m-j)*n*j;//6
25                 tmp=tmp+(double)n*m;//5
26                 tmp=tmp+(double)(j-1)*n*(m-j+1);//4
27                 tmp=tmp+(double)(n-i)*(j-1)*i*(m-j+1);//7
28                 tmp=tmp+(double)(n-i)*i*m;//8
29                 tmp=tmp+(double)(n-i)*(m-j)*i*j;//9
30
31                 double p=tmp/n/n/m/m;
32                 ans=ans+1-pow((1-p),k);
33
34             }
35         }
36         printf("Case #%d: ",++ac);
37         printf("%d\n",int(ans+0.5));
38     }
39     return 0;
40 }