hdu 3664Permutation Counting(简单dp)

Permutation CountingTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1175    Accepted Submission(s): 589


Problem DescriptionGiven 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. 
InputThere are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N). 
 
OutputOutput 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 03 1

 
Sample Output

14HintThere 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} 

 
Source2010 Asia Regional Harbin 

题目大意：给你一个n，代表有n个数，分别是1 2 3 .......n-1 n，如果这个数比这个数的位置大，那么t++,例如3个数，1 3 2那么t=1,因为a[2]>2，给你一个n，再给你一个k，问你这样使得t等于k的排列组合数是多少？
 解题思路：开始是自己写了前面四个，可以慢慢发现规律，设比它位置大的数为特殊数，f[i][j] 表示从1到i 特殊数为j 的排列数

(1) f[i-1][j]表示添加的第i 个数在第 i个位置,不改变特殊数的个数

(2) j*f[i-1][j] 将第i 个数和前面j 个特殊数数交换位置,依然不改变特殊数的个数

(3) (i-j)*f[i-1][j-1] 将第i 个数和前面 (i-j) 个非特殊数数交换位置可得到j 个特殊数

具体实现见代码。

题目地址：Permutation Counting

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;long long mod = 1e9+7;long long ans[1002][1002];void presolve(){    int i,j;    for(i=1;i<=1000;i++)    {        ans[i][0]=1;        for(j=1;j<=1000;j++)            ans[i][j]=((1+j)*ans[i-1][j]+(i-j)*ans[i-1][j-1])%mod;    }}int main(){    presolve();    int n,k;    while(cin>>n>>k)        cout<<ans[n][k]<<endl;    return 0;}