HDU 5245 Joyful (概率题 求期望)

Joyful

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

[b]Total Submission(s): 478 Accepted Submission(s): 209[/b]

Problem Description
Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like anM×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 theM×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 integerT(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 thet-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.

Source
The 2015 ACM-ICPC
China Shanghai Metropolitan Programming Contest

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5245

题目大意：一个大矩形由n*m个小方块组成，一个人每次选两个小方块，他可以对以这两个方块为顶点的矩形面积内的小方块上色，他一共可选k次，现在求被上色的小方块数目的期望

题目分析：这是最伤心的一题，上海大都会邀请赛，出了这题就拿银牌了，无奈坑了队友

每个格子可以被重复选，因此可以把每一个小方块选不选当做一个独立事件，所以我们算出每个小方块对总期望的贡献值就行了，直接算不好算，考虑算每个小方块一次不被选的概率p，这个算起来就方便很多，不妨以当前点为中心，那么只有四种情况，被选的两个小方块同在中心的上下左右，不过这样会重复，四个角相当于被算了两次，再把它们减掉一次，这样求出来的是当前点一次不会被上色的情况数，除以总情况数(m * n * m * n)就是当前方块一次不会被上色的概率，再乘k次，就是k次这个方块都不被上色的概率，用1减则为选k次这个方块被上色的概率，只需要把每个方块选k次后被上色的概率累加即可，情况数中间可能会爆int，用long
long

#include <cstdio>
#include <cmath>
#define ll long long

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        int m, n, k;
        scanf("%d %d %d", &m, &n, &k);
        double ans = 0;
        ll sum = (ll) m * m * n * n;
        for(int i = 1; i <= m; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                ll num = 0;
                num += (ll) (i - 1) * (i - 1) * n * n;
                num += (ll) (m - i) * (m - i) * n * n;
                num += (ll) (j - 1) * (j - 1) * m * m;
                num += (ll) (n - j) * (n - j) * m * m;
                num -= (ll) (i - 1) * (i - 1) * (j - 1) * (j - 1);
                num -= (ll) (i - 1) * (i - 1) * (n - j) * (n - j);
                num -= (ll) (j - 1) * (j - 1) * (m - i) * (m - i);
                num -= (ll) (m - i) * (m - i) * (n - j) * (n - j);
                double p = 1.0 * num / sum;
                p = pow(p, k);
                ans += 1.0 - p;
            }
        }
        printf("Case #%d: %d\n", ca, (int)round(ans));
    }
}