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  • ACM HDU 3664 Permutation Counting

    Permutation Counting

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 493    Accepted Submission(s): 258


    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
     

    Recommend
    lcy
     

    题意:对于任一种N的排列A,定义它的E值为序列中满足A[i]>i的数的个数。给定N和K(K<=N<=1000),问N的排列中E值为K的个数。

    解法:简单DP。dp[i][j]表示i个数的排列中E值为j的个数。假设现在已有一个E值为j的i的排列,对于新加入的一个数i+1,将其加入排列的方法有三:1)把它放最后,加入后E值不变    2)把它和一个满足A[k]>k的数交换,交换后E值不变       3)把它和一个不满足A[k]>k的数交换,交换后E值+1      根据这三种方法得到转移方程dp[i][j] = dp[i - 1][j] + dp[i - 1][j] * j + dp[i - 1][j - 1] * (i - j);

    #include<stdio.h>
    #include
    <iostream>
    using namespace std;
    const int MAXN=1001;
    long long dp[MAXN][MAXN];
    const long long MOD=1000000007;
    int main()
    {
    int n,k;
    int i,j;
    for(i=1;i<=1000;i++)
    {
    dp[i][
    0]=1;
    for(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)!=EOF)
    printf(
    "%I64d\n",dp[n][k]);
    return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2119660.html
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