HDU 5245 Joyful (概率题 求期望)

Joyful

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

Total Submission(s): 478    Accepted Submission(s): 209

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

23 3 14 4 2

 

Sample Output

Case #1: 4Case #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 longint 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));    }}