HDU 1028 Ignatius and the Princess III

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

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

41020

Sample Output

542627

这道题就是将从1到n的所有数当作背包要装的物品，n为背包容量来做完全背包。

因为后面的肯定要比前面的大，所以状态转移方程可以写为后面直接加上前面的。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int inf=999999;int main(){    int n;    int res[150];    while(cin>>n)    {        memset(res,0,sizeof(res));        res[0]=1;        for(int i=1;i<=n;i++)        {            for(int j=i;j<=n;j++)            {                res[j]+=res[j-i];            }        }        cout<<res[n]<<endl;    }    return 0;}