Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16742    Accepted Submission(s): 11783


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 4 10 20

//
//  main.cpp
//  數的拆分（母函式的應用）
//
//  Created by 張嘉韜 on 16/1/26.
//  Copyright © 2016年 張嘉韜. All rights reserved.
//

#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
    //freopen("")
    int n,c1[200],c2[200];
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<=n;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(int i=2;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                for(int k=0;j+k<=n;k=k+i)
                {
                    c2[k+j]+=c1[j];
                }
            }
            for(int j=0;j<=n;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        cout<<c1[n]<<endl;
    }
    return 0;
}
 

母函式解析http://blog.csdn.net/vsooda/article/details/7975485