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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.
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.
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.
2 3 3 1 4 4 2
Case #1: 4 Case #2: 8
The precise answer in the first test case is about 3.56790123.
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 }
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原文地址：http://www.cnblogs.com/UniqueColor/p/4792663.html
