hdu 1028 Ignatius and the Princess III

Ignatius and the Princess III

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

Total Submission(s) : 18   Accepted Submission(s) : 13

Font: Times New Roman | Verdana | Georgia 

Font Size: ← →

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

#include <stdio.h>int main(){    int c1[10001],c2[10001];    int  i,j,k,n;    while (scanf("%d",&n)!=EOF)    {            for (i=0;i<=n;i++)                    {    c1[i]=0;c2[i]=0;}        for (i=0;i<=n;i++)    c1[i]=1;        for(i=2;i<=n;i++)        {            for(j=0;j<=n;j++)                for(k=0;j+k<=n;k=k+i)                    c2[j+k]+=c1[j];        for (j=0;j<=n;j++)        {            c1[j]=c2[j];            c2[j]=0;        }        }            printf("%d\n",c1[n]);    }        return 0;}