hdu 5245 Joyful（期望）

Joyful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1243    Accepted Submission(s): 546

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

Hint

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

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

Recommend

直接求好像不大好求，可以反着求

代码较挫，

1 #include

2 #include

3 using namespace std;

4

5 using namespace std;

6

7 #define LL long long

8

9 long long C(LL a,LL b)

10 {

11 return a * b * a * b;

12 }

13 int main()

14 {

15 long long T,m,n,k;

16 int cas = 0;

17 scanf("%lld",&T);

18 while(T--)

19 {

20 LL sum;

21 double ans=0, p;

22 scanf("%lld%lld%lld",&m,&n,&k);

23 for(LL i=0; i

24 {

25 for(LL j=0; j

26 {

27 sum = C(i,n)+C((m-i-1),n)+C(j,m)+C((n-j-1),m);

28 sum = sum - C(i,j) - C(i,(n-j-1)) - C((m-i-1),j ) - C((m-i-1),(n-j-1)) ;

29 p = sum * 1.0 / C(m,n);

30

31 double tmp=1;

32 for (LL jj = 1; jj <= k; ++jj)

33 tmp *= p;

34 ans=ans+1-tmp;

35 }

36 }

37 printf("Case #%d: %.0f\n",++cas,ans);

38 }

39 return 0;

40 }

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