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 1st position and 3 in 3rd 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：先从m个数中选出k个数待在自己位置上（c(m, k)）， 然后在剩下n-k个数中，有n-m个数可以待在自己原来的位置上，故每局n-m个数有多少个数在自己原来的位置上，剩下的 n - k - i 个数 就是一个错排列了。
（错排列 ： D[i] = (i-1) * (D[i-1] + D[i-2] ) ; ）
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1100
#define INF 0x3f3f3f3f
#define mod 1000000007
long long c[N][N], d[N];
void init()///组合数c(n, m)的的值存在c数组中。
{
c[0][0] = 1;
for(int i = 1; i < N; i++)
{
c[i][0] = c[i][i] = 1;
for(int j = 1; j < i; j++)
{
c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod;
}
}
d[1] = 0;
d[2] = d[0] = 1;
for(int i = 3; i < N; i++)
d[i] = (i-1) * (d[i-1] + d[i-2]) % mod;///错排列 ： D[i] = (i-1) * (D[i-1] + D[i-2] )
}
long long solve(int n, int m, int k)
{
long long ans = 0;
for(int i = 0; i <= n - m; i++)///i表示在n-m数中选i个数呆在自己原来的位置。
ans = (ans + (c[n-m][i] * d[n-k-i]) % mod) % mod;///c[n-m][i] * d[n-k-i]表示在n-m数中选i个数呆在自己原来的位置，剩下n-k-i都不呆在自己位上的数错排列得到的排列数。
return ans * c[m][k] % mod;
}
int main()
{
int T, cas;
int n, m, k;
scanf("%d", &T);
cas = 0;
init();
while(T--)
{
cas++;
scanf("%d%d%d", &n, &m, &k);
long long ans = solve(n, m, k);
printf("Case %d: %lld\n", cas, ans);
}
return 0;
}