DP题目 sicily 1687 Permutation

DP 水题

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

66

当前状态：dp[i][j]表示i个小于号，j个数字
可以转移到当前状态的有：dp[i][j-1]*(1+i)+dp[i-1][j-1]*(j-i)
初始： dp[0][i]=1;
       1<2<4>3   
       再加一个数字5时，5可以插入的位置有4种：最左 最右 <  >
       插入后对 < 的影响：
            最左 :  不变
            最右 ： +1
              <  :   ....  
              >  :   ....
       然后统计下有多少个<>~~
分析够彻底了。供我们这些小菜鸟参考应该还行~~
这题水分还是足够分量的。

 if(j>=i)
       dp[i][j]=dp[i][j-1]*(1+i)+dp[i-1][j-1]*(j-i);
  if(dp[i][j]>=2007)          因为%的复杂度高，所以不放在上面。。当然也是个人习惯
        dp[i][j]=dp[i][j]%2007;
              //cout<<"1 "<<dp[i][j]<<endl;     可以观察每一步怎么走的。。

#include<cstdio>
int dp[200][200];    //[i][j]表示i个小于号，j个数字
int n,k;

int main()
{
    while(scanf ("%d%d",&n,&k)!=EOF){
        for(int i=0;i<=n;++i)
            dp[0][i]=1;
        for(int j=1;j<=n;++j)
           for(int i=1;i<=k;++i)
            {
                if(j>=i)
                    dp[i][j]=dp[i][j-1]*(1+i)+dp[i-1][j-1]*(j-i); 
                if(dp[i][j]>=2007)
                    dp[i][j]=dp[i][j]%2007;
                    //cout<<"1 "<<dp[i][j]<<endl;
            }   
        printf("%d
",dp[k][n]);
    }
    return 0;
}