（专题DP）⁽⁽ ◟(눈_눈)◞ ⁾⁾

题意：几个字母比大小的组合数，注意相同字母还有符号

题解：动态规划

Description
Background

Consider a specific set of comparable objects. Between two objects a and b, there exits one of the following three classified relations:
a = b
a < b
b < a
Because relation '=' is symmetric, it is not repeated above.
So, with 3 objects ( a, b, c), there can exist 13 classified relations:
a = b = c a = b < c c < a = b a < b = c
b = c < a a = c < b b < a = c a < b < c
a < c < b b < a < c b < c < a c < a < b
c < b < a
Problem

Given N, determine the number of different classified relations between N objects.
Input
Includes many integers N (in the range from 2 to 10), each number on one line. Ends with −1.
Output
For each N of input, print the number of classified relations found, each number on one line.
Sample Input
input output
2
3
-1
3
13

#include<stdio.h>
#include<string.h>
#define N 10
int main()
{
    int dp[N][N],a[N];
    int n,i,t,j;
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        dp[1][1]=1;
        for(i=2;i<N;i++)//i代表有几个字母 
        {
            a[i]=0;
            for(j=1;j<N;j++)//j代表有几个不同字母 
            {
                dp[i][j]=dp[i-1][j-1]*j+dp[i-1][j]*j;
                           //加一个小于号 //加一个等号 
                a[i]+=dp[i][j];
            }
        }
    while(~scanf("%d",&n))
    {
        if(n==-1)
            break;
        printf("%d
",a[n]);
    }
    return 0;
}