Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10143    Accepted Submission(s): 7184

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

 

#include<iostream>
#include<cstring>
using namespace std;
int num[130];
int dp[130][130];
int main()
{
    int n;
    int i,j,k;
    while(cin>>n)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(i=1;i<=n;i++)
        {
            num[i]=i;
        }
        for(i=1;i<=n;i++)
        {
            for(j=0;j<=n;j++)
            {
                dp[i][j]=dp[i-1][j];
                for(k=1;k*num[i]<=j;k++)
                {
                    dp[i][j]=dp[i][j]+dp[i-1][j-k*num[i]];
                }
            }
        }
        cout<<dp[n][n]<<endl;
    }
    return 0;
}