Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Sample Output

5
42
627

Author

Ignatius.L

题目不能再经典了！简直就是 模版的雏形！

#include<iostream>
using namespace std;
int main()
{
    int n,i,j,k,cnt[125],dic[125];
    while(cin>>n)
    {
        for(i=0; i<=n; i++)
        {
            cnt[i]=1;
            dic[i]=0;
        }

        for( int m=2; m<=n; m++)
        {

            for(j=0; j<=n; j++)
                for(k=0; k+j<=n; k+=m)
                {
                    dic[j+k]+=cnt[j];

                }
            for(int p=0; p<=n; p++)
            {

                cnt[p]=dic[p];
                dic[p]=0;
            }
        }
        cout<<cnt[n]<<endl;
    }
}