HDU 3664 Permutation Counting

Permutation Counting

http://acm.hdu.edu.cn/showproblem.php?pid=3664

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1001    Accepted Submission(s): 496

Problem Description Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations of {1, 2, …, N} whose E-value is exactly k.   Input There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).   Output Output one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.   Sample Input   3 0 3 1   Sample Output   1 4 Hint There is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}   Source 2010 Asia Regional Harbin   題意：求a[i]>i 的個數。。。。DP就ok了，，dp[i][j]表示為i個數排列E為j的個數   将第i個數插入到數組中，有三種情況： 　　①插入到末尾，則個數保持不變，為dp[i-1][j]; 　　②與a[i]>i的交換，則個數亦保持不變,但有j中情況，共為dp[i-1][j]*j; 　　③與a[i]<i的交換，則個數增加1，有i-j種情況，共為dp[i-1][j-1]*(i-j);   狀态轉移方程：dp[i][j]=dp[i-1][j]+dp[i-1][j]*j+dp[i-1][j-1]*(i-j);    

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int mod=1000000007;

int n,k;
long long dp[1010][1010];   //注意精度。。。dp[i][j]表示為i個數排列E為j的個數

int main(){

    //freopen("input.txt","r",stdin);

    for(int i=1;i<=1000;i++){   //打表，否則逾時
        dp[i][0]=1;
        for(int j=1;j<=i;j++)
            dp[i][j]=(dp[i-1][j]+dp[i-1][j]*j+dp[i-1][j-1]*(i-j))%mod;
    }
    while(~scanf("%d%d",&n,&k)){
        cout<<dp[n][k]<<endl;
    }
    return 0;
}