HDU 1028 Ignatius and the Princess III

Problem Description

"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

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

 

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

Sample Input

41020

 

Sample Output

542627
经典DP
#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;    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 33;}