"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!"

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.

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

Ignatius.L

 

#include <iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int main()
{
    int c1[150],c2[150],n;
    for(int i=0; i<=130; i++)
        c1[i]=1,c2[i]=0;
    for(int i=2; i<=130; i++)
    {
        for(int j=0; j<=130; j++)
            for(int k=0; k+j<=130; k+=i)
                c2[k+j]+=c1[j];
        for(int j=0; j<=130; j++)
            c1[j]=c2[j],c2[j]=0;
    }
    while(cin>>n)
        cout<<c1[n]<<endl;

    return 0;
}