hdu1028 Ignatius and the Princess III (完全背包)

Ignatius and the Princess III

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

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Total Submission(s): 16474    Accepted Submission(s): 11609

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

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代码：

#include<cstdio>
#include<cstring>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;

const int maxn=120;
int f[maxn+10];

int main()
{
  freopen("1.in","r",stdin);
  
  int n,i,j;
  ms(f),f[0]=1;
  for(i=1;i<=maxn;i++)
    for(j=i;j<=maxn;j++)f[j]+=f[j-i];  
  while(scanf("%d",&n)==1)printf("%d\n",f[n]);
  return 0;
}