1687. Permutation_permutation plays a very important role in combina-优快云博客

本文探讨了如何计算给定排列中特定符号的数量，并通过实例展示了计算过程，包括输入和输出样例。

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1687. Permutation

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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 symbols '<' or '>' between every pairs of consecutive numbers of a permutation, we can get the permutation 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 your task to calculate the number of n-permutations with k '<' symbols.

Maybe you don't like large numbers, so you should just give the result mod 2007.

Input

Input may contain 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 nonnegative integer result mod 2007 on a line.

Sample Input

5 2

Sample Output

何叶红同学的解答非常好，大家鼓掌

点击打开链接

代码如下：

#include <iostream>
#include <cstring>
#include <stdio.h> // for scanf and EOF
using namespace std;

int dp[105][105];
void init()
{
    dp[0][0]=1;
    for (int i = 1; i < 105; ++i)
    {
        /* code */
        dp[i][0]=1;
        for (int j=1;j<i-1;j++)
        {
            /* code */
            dp[i][j]=(dp[i-1][j]*(j+1)%2007+dp[i-1][j-1]*(i-j)%2007)%2007;
        }
        dp[i][i-1]=1;
    }
}
int main()
{
 init();
 int a,b;
 while(scanf("%d%d",&a,&b)!=EOF)
 {
    cout<<dp[a][b]<<endl;

 }
  
   // system("pause");
}