C - Cheerleaders

In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Their roles are substantial during breaks and prior to start of play. The world cup soccer is no exception. Usually the cheerleaders form a group and perform at the centre of the field. In addition to this group, some of them are placed outside the side line so they are closer to the spectators. The organizers would like to ensure that at least one cheerleader is located on each of the four sides. For this problem, we will model the playing ground as an M*N rectangular grid. The constraints for placing cheerleaders are described below:

 

§  There should be at least one cheerleader on each of the four sides. Note that, placing a cheerleader on a corner cell would cover two sides simultaneously.

§  There can be at most one cheerleader in a cell.

§  All the cheerleaders available must be assigned to a cell. That is, none of them can be left out.

 

 

The organizers would like to know, how many ways they can place the cheerleaders while maintaining the above constraints. Two placements are different, if there is at least one cell which contains a cheerleader in one of the placement but not in the other. 

 

 

 

Input

 

The first line of input contains a positive integer T<=50, which denotes the number of test cases. T lines then follow each describing one test case. Each case consists of three nonnegative integers, 2<=M, N<=20 and K<=500. Here M is the number of rows and N is the number of columns in the grid. K denotes the number of cheerleaders that must be assigned to the cells in the grid.

 

 

Output

 

For each case of input, there will be one line of output. It will first contain the case number followed by the number of ways to place the cheerleaders as described earlier. Look at the sample output for exact formatting. Note that, the numbers can be arbitrarily large. Therefore you must output the answers modulo1000007.

 

Sample Input

Sample Output

2

2 2 1

2 3 2

Case 1: 0

Case 2: 2

题意：给一个m*n的方格在上面放k个物体，要求第一行第一列最后一行最后一列都必须存在物体，求有多少种方案。

思路：本质上就是在m*n个人中选择k个人的方案数，然后加了一个要求。类似于数学中的组合数，可以用杨辉三角打表。

   容斥原理加二进制枚举

设 A 为第一行没人     

B为最后一行没人

C为第一列没人

D为最后一列没人

那么我们想要的结果      M=全集-A-B-C-D+AB+AC+AD+BC+BD+CD-BCD-ACD-ABD-ABC+ABCD

代码中使用1~15的二进制把上述16种情况集合表示了。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MOD=1000007;
const int N=540;
int C[N][N];
void zuheshu()
{
    memset(C,0,sizeof(C));
    C[0][0]=1;


    for(int i=0; i<=500; i++)
    {
        C[i][0]=C[i][i]=1;
        for(int j=1; j<i; j++)
        {
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;

        }

    }
}

int main()
{
    zuheshu();
    int t;
    cin>>t;
    for(int j=1; j<=t; j++)
    {
        int m,n,k,r,c,b;
        cin>>n>>m>>k;
        int ans=0;
        for(int i=0; i<16; i++) //用0001代表A，0010代表B，0100代表C，1000代表D
        {
            b=0;
            r=n;
            c=m;
            if(i&1)
            {
                r--;    //判断第0位是否为1即第一行没有人。
                b++;
            }
            if(i&2)
            {
                r--;    //判断第1位是否为1
                b++;
            }
            if(i&4)
            {
                c--;    //判断第2位是否为1
                b++;
            }
            if(i&8)
            {
                c--;    //判断第3位是否为1
                b++;
            }
            if(b&1)//如果b为奇数，表示减去，若为偶数表示加上。 结果=全集-A-B-C-D+AB+AC+AD+BC+BD+CD-BCD-ACD-ABD-ABC+ABCD
            {
                ans=(ans+MOD-C[r*c][k])%MOD;
            }
            else
                ans=(ans+C[r*c][k])%MOD;
        }
        printf("Case %d: %d\n",j,ans);
    }



}