hdu 2583 permutation 好题递推

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本文介绍了一种计算特定符号排列数量的方法，通过递推公式计算n个数的全排列中有k个“小于”符号的排列数量，并给出了解决方案的代码实现。

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permutation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 121 Accepted Submission(s): 67

Problem Description

Permutation plays a very important role in Combinatorics. For example ,1 2 3 4 5 and 1 3 5 4 2 are both 5-permutations. As everyone's known, the number of n-permutations is n!. According to their magnitude relatives ,if we insert the sumbols "<" or ">"between every pairs of consecutive numbers of a permutations,we can get the permutations with symbols. For example,1 2 3 4 5 can be changed to 1<2<3<4<5, 1 3 5 4 2 can be changed to 1<3<5>4>2. Now it's yout task to calculate the number of n-permutations with k"<"symbol. Maybe you don't like large numbers ,so you should just geve the result mod 2009.

Input

Input may contai multiple test cases.
Each test case is a line contains two integers n and k .0<n<=100 and 0<=k<=100.
The input will terminated by EOF.

Output

The nonegative integer result mod 2007 on a line.

Sample Input

5 2

Sample Output

Source

ECJTU 2009 Spring Contest

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题意：输入 n m 问n组成的全排列中有m个小于号的排列的个数

思路：

我们设

n（<=100）个数 1,2,...,n的置换有n!个，有k个小于号的有 f[n][k]个。

假如

a[1],a[2],...,a[n-1],是n-1个数的置换，有 k个'<'

把 n插入，有n个位置可以插入,
把 n插入到a[1]的前面, '<'没有增加 , 1个位置
如 a[i]<a[i+1], 把 n插入到a[i+1]的前面, '<'没有增加 , k个位置

插入到其他位置 '<'增加
因此
f[n][k]=f[n-1][k]*(k+1) + f[n-1][k-1] *(n-k)

思路来自：http://hi.baidu.com/fhhgoeqlilqrtxr/item/4c8c0b768905a7336dc37cc4

#include<stdio.h>
int ans[111][111];
int main()
{
    int n,k,i,j;
    for(i=0;i<=100;i++)
    {
        ans[i][0]=1;
        ans[0][i]=0;
    }
    for(i=1;i<=100;i++)
        for(j=1;j<=100;j++)
           {
               ans[i][j]=ans[i-1][j]*(j+1)+ans[i-1][j-1]*(i-j);
               ans[i][j]%=2009;
           }
    while(scanf("%d %d",&n,&k)!=EOF)
    {
             printf("%d\n",ans[n][k]);
    }
    return 0;
}