递归。。。

#include<stdio.h>
#include<string.h>
const int MAXN=130;
int dp[MAXN][MAXN];
int calc(int n,int m)
{
   
    if(dp[n][m]!=-1) return dp[n][m];
   
   
    if(n<1||m<1) return dp[n][m]=0;
    if(n==1||m==1) return dp[n][m]=1;
    if(n<m) return dp[n][m]=calc(n,n);
    if(n==m) return dp[n][m]=calc(n,m-1)+1;
 else
    return dp[n][m]=calc(n,m-1)+calc(n-m,m);
   
}    
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
   
    while(scanf("%d",&n)!=EOF)
      printf("%d\n",calc(n,n));
    return 0;
}

Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input

Output

Sample Input

4
10
20

Sample Output

5
42
627

递归。。。,布布扣,bubuko.com

递归。。。

原文：http://www.cnblogs.com/luzhongshan/p/3892512.html