HDOJ5245 Joyful(dp)

Joyful

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 506 Accepted Submission(s): 219



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

HintThe precise answer in the first test case is about 3.56790123.

概率dp，谁做出来商kalili香吻一枚2333

参考了点击打开链接

就是把矩阵拆分，对每个小矩阵计算概率。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "cmath"
using namespace std;
int t, k;
double n, m;
int main(int argc, char const *argv[])
{
	scanf("%d", &t);
	for(int cas = 1; cas <= t; ++cas) {
		scanf("%lf%lf%d", &n, &m, &k);
		double ans = 0;
		for(double i = 1; i <= n; ++i)
			for(double j = 1; j <= m; ++j) {
				double p = m * n;
				p += (i - 1) * (j - 1) * (n - i + 1) * (m - j + 1);
				p += (i - 1) * (m - j) * (n - i + 1) * j;
				p += (j - 1) * (n - i) * (m - j + 1) * i;
				p += (n - i) * (m - j) * i * j;
				p += (i - 1) * m * (n - i + 1);
				p += (m - j) * n * j;
				p += (n - i) * m * i;
				p += (j - 1) * n * (m - j + 1);
				p = p / n / n / m / m;
				ans += 1 - pow(1 - p, k);
			}
		printf("Case #%d: %d\n", cas, int(ans + 0.5));
	}
	return 0;
}