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≤), 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 ≤M,N≤, ≤K≤.

 

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

 

Sample Output

Case #:
Case #: 

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、、、9四个区域内任取;
②[x1,y1]取在区域2内时，[x2,y2]可以在4、、、、、9六个区域内任取;
③[x1,y1]取在区域3内时，[x2,y2]可以在4、、、8四个区域内任取;
④[x1,y1]取在区域4内时，[x2,y2]可以在2、、、、、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、、、、、8六个区域内任取;
⑦[x1,y1]取在区域7内时，[x2,y2]可以在2、、、6四个区域内任取;
⑧[x1,y1]取在区域8内时，[x2,y2]可以在1、、、、、6六个区域内任取;
⑨[x1,y1]取在区域1内时，[x2,y2]可以在1、、、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

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<stdlib.h>
 #include<queue>
 using namespace std;
 #define ll long long
 int m,n,k;
 int main()
 {
     int t;
     int ac=;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d%d",&n,&m,&k);
         double ans=;
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 double tmp=;
                 tmp=tmp+(double)(i-)*(j-)*(n-i+)*(m-j+);//
                 tmp=tmp+(double)(i-)*(n-i+)*m;//
                 tmp=tmp+(double)(i-)*(m-j)*(n-i+)*j;//
                 tmp=tmp+(double)(m-j)*n*j;//
                 tmp=tmp+(double)n*m;//
                 tmp=tmp+(double)(j-)*n*(m-j+);//
                 tmp=tmp+(double)(n-i)*(j-)*i*(m-j+);//
                 tmp=tmp+(double)(n-i)*i*m;//
                 tmp=tmp+(double)(n-i)*(m-j)*i*j;//

                 double p=tmp/n/n/m/m;
                 ans=ans+-pow((-p),k);

             }
         }
         printf("Case #%d: ",++ac);
         printf("%d\n",int(ans+0.5));
     }
     return ;
 }