【DP】hdu 3664

#include <list>#include <map>#include <set>#include <queue>#include <string>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>#include <string.h>using namespace std;#define LL long long#define PI acos(-1.0)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-10#define FRE freopen("a.txt","r",stdin)#define N 1005#define MOD 1000000007/*给出1……n的排列,求解满足a[i]>i的个数为K的排列的个数.用dp[n][k]表示1……n的排列里面有k个a[i]>i,现在考虑第n个数,即a[n]=n    1> 如果a[n]=n与前面的任何一个a[i]<=i交换,则满足a[i]>i的位置数加1,所以前面满足a[i]>i的位置数为k-1(即排列总数为dp[n-1][k-1]),a[n]前面满足a[i]<=i个位置共有(n-1-(k-1))=(n-k),故dp[n][k]=(n-k)*dp[n-1][k-1]    2> 如果a[n]=n与前面的任何一个a[i]>i交换或者a[n]=n时,则满足a[i]>i的位置数不变,所以前面满足a[i]>i的位置数为k(即排列总数为dp[n-1][k]),则dp[n][k]=(k+1)*dp[n-1][k]综上,dp[n][k]=(n-k)*dp[n-1][k-1]+(k+1)*dp[n-1][k]*/LL dp[N][N];int main(){    int i,j,k;    dp[0][0]=1;    for(i=1;i<=1000;i++){        for(j=0;j<=i;j++){            dp[i][j]=(j+1)*dp[i-1][j]%MOD;            if(j)            dp[i][j]=(dp[i][j]+(i-j)*dp[i-1][j-1])%MOD;        }    }    int n;    while(scanf("%d%d",&n,&k)!=EOF){        printf("%I64d\n",dp[n][k]);    }    return 0;}