Ignatius and the Princess III

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

Total Submission(s): 16122    Accepted Submission(s): 11371

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<stdio.h>
#include<string.h>
int dp[1000][1000],n;
int main()
{
while(scanf("%d",&n)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<1000;i++)
dp[i][1]=dp[1][i]=1;
for(int i=2;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i>j) dp[i][j]=dp[i-j][j]+dp[i][j-1];
else if(i==j) dp[i][j]=dp[i][j-1]+1;
else dp[i][j]=dp[i][i];
}
}
printf("%d\n",dp[n][n]);
}
return 0;
}*/
#include<stdio.h>
#include<string.h>
int dp[1200];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(dp,0,sizeof(dp));
dp[0]=1;
for(int i=1;i<=n;i++)
for(int j=i;j<=n;j++)
{
dp[j]+=dp[j-i];
}
printf("%d\n",dp[n]);
}
}

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