Light OJ 1095 Arrange the Numbers(計數）

Consider this sequence {1, 2, 3 ... N}, as an initial sequence of first N natural numbers. You can rearrange this sequence in many ways. There will be a total of N! arrangements. You have to calculate the number of arrangement of first N natural numbers, where in first M positions; exactly K numbers are in their initial position.

For Example, N = 5, M = 3, K = 2

You should count this arrangement {1, 4, 3, 2, 5}, here in first 3 positions 1 is in 1^st position and 3 in 3^rd position. So exactly 2 of its first 3 are in there initial position.

But you should not count {1, 2, 3, 4, 5}.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case contains three integers N (1 ≤ N ≤ 1000), M (M ≤ N), K (0 < K ≤ M).

Output

For each case, print the case number and the total number of possible arrangements modulo 1000000007.

Sample Input	Output for Sample Input
2 5 3 2 10 6 3	Case 1: 12 Case 2: 64320

 

分析：K個數位置不變，剩餘m-k個數肯定參與錯排，那麼n-m個數裏參與錯排的個數枚舉下就可以算出來了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1000000007;
LL dp[1010],C[1010][1010];
void init()
{
    C[1][1]=C[1][0]=C[0][0]=1;
    for(int i=2;i<=1000;i++)
    {
        C[i][i]=C[i][0]=1;
        for(int j=1;j<i;j++)
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
    dp[1]=0;dp[2]=1;dp[0]=1;
    for(int i=3;i<=1000;i++)
        dp[i]=1LL*(i-1)*(dp[i-1]+dp[i-2])%mod;
}
LL Cal(int n,int m,int k)
{
    int x=n-m;
    int y=m-k;
    LL ans=0;
    for(int i=0;i<=x;i++)
       ans=(ans+C[x][i]*dp[y+i]%mod)%mod;
    return ans*C[m][k]%mod;
}
int main()
{
    init();
    int t,cas=1;
    int n,m,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        printf("Case %d: %lld\n",cas++,Cal(n,m,k));
    }
    return 0;
}