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): 6541    Accepted Submission(s): 4620

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>
#define MAX 120

int dp[MAX+5][MAX+5];

int ans(int n,int max){
	if(dp[n][max])
		return dp[n][max];
	if(n<=max){
		if(!dp[n][n-1])
			dp[n][n-1]=ans(n,n-1);
		return dp[n][n-1]+1;
	}
	if(!dp[n][max-1])
		dp[n][max-1]=ans(n,max-1);
	if(!dp[n-max][max])
		dp[n-max][max]=ans(n-max,max);
	return dp[n][max-1]+dp[n-max][max];
}

int main(){
	int n,i;
	memset(dp,0,sizeof(dp));
	for(i=1;i<=MAX;i++)
		dp[i][1]=dp[1][i]=1;
	for(i=2;i<=MAX;i++)
		dp[i][i]=ans(i,i);
	while(~scanf("%d",&n))
		printf("%d\n",dp[n][n]);
	return 0;
}