HDU 1028 整数拆分问题 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): 15498    Accepted Submission(s): 10926

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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#include<stdio.h>
#include<string.h>
int ans[],temp[];
int main(){
  int n;
  while(scanf("%d",&n)!=EOF){
      memset(ans,,sizeof(ans));
      memset(temp,,sizeof(temp));
      for(int i=;i<=n;i++)
        ans[i]=;
      for(int i=;i<=n;i++){
          for(int j=;j<=n;j++){
              for(int k=;k+j<=n;k+=i){
                   temp[k+j]+=ans[j];
              }
          }
        for(int ti=;ti<=n;ti++){
            ans[ti]=temp[ti];
            temp[ti]=;
        }

      }

    printf("%d\n",ans[n]);
  }
  return ;
}