算法-dp The Staircases

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:
Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.
Input
Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.
Output
Numbers Q, one on each line.
Sample Input
3
5
0
Sample Output
1
2

#include
#include
#include
#include
#include
using namespace std;
long long dp[510];
int main()
{
	memset(dp,0,sizeof(dp));
	dp[0]=1;
	for(int i=1;i<=500;i++){
		for(int j=500;j>=i;j--){
			dp[j]+=dp[j-i]; //在最后插上一个高度为i的砖头列 
			//j为容量，即共使用多少块砖头
		}
	}
	int n;
	while(~scanf("%d",&n))
	{
		if(n==0) break;
		printf("%lld\n",dp[n]-1);//减去只有一列为n块的情况
	}
	return 0;
}
/*
3
5
0 
*/