hdu 1028 Ignatius and the Princess III(DP)

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

::简单的DP题

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
ll dp[125]={1,0};

int main()
{
    ios::sync_with_stdio(0);
    for(int i=1; i<=120 ; i++ )//当只有不大于i的数
        for(int j =i ; j<=120 ; j++ )
            dp[j] += dp[j-i];

    int n;
    while(cin>>n)
        cout<<dp[n]<<endl;
    return 0;
}