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  • 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
     
    题意:求a[i]>i 的个数。。。。DP就ok了,,dp[i][j]表示为i个数排列E为j的个数
     
    将第i个数插入到数组中,有三种情况:
      ①插入到末尾,则个数保持不变,为dp[i-1][j];
      ②与a[i]>i的交换,则个数亦保持不变,但有j中情况,共为dp[i-1][j]*j;
      ③与a[i]<i的交换,则个数增加1,有i-j种情况,共为dp[i-1][j-1]*(i-j);
     
    状态转移方程: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]表示为i个数排列E为j的个数
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        for(int i=1;i<=1000;i++){   //打表,否则超时
            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;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/2839379.html
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