hdu 1028 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): 13312    Accepted Submission(s): 9420

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

 

Author

Ignatius.L

 

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#include<stdio.h>
#include<string.h>
int main()
{
int c2[200]={0},c1[200];
int i,j,k,n;
for(i=0;i<200;i++)
c1[i]=1;
for(i=2;i<130;i++)
{
for(j=0;j<=120;j++)
{
for(k=0;k+j<=120;k+=i)
{
c2[k+j]+=c1[j];
}
}
for(j=0;j<=120;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",c1[n]);
}
return 0;
}