*Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7012 Accepted Submission(s): 4968
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 */
#include<iostream>
#include<stdio.h>
using namespace std;
//const int max=10001;//注意不能用这个
#define max 10000000
int c1[max];
int c2[max];
int main()
{
int i,j,k,n;
while(scanf("%d",&n)!=EOF)
{
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+=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;
}