Permutation Counting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1794 Accepted Submission(s): 955
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}
和上一篇csu多校的差不多,替换i个位置(最后一个位置直接放):
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。
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);
注意i和j从1开始,maxn是不能等到,越界
#include <bits/stdc++.h> using namespace std; const int maxn=1000+10; long long dp[maxn][maxn]; const int mod = 1000000007; int main() { int n,k; int i,j; for(i=1;i<maxn;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)) printf("%I64d\n",dp[n][k]); return 0; }