题解链接：

https://acmer.blog.csdn.net/article/details/123722816

视频链接：

https://www.bilibili.com/video/BV1bY4y1s7J3/

动态规划：dp[i][j]表示前i个砝码能不能称出j的重量，最后统计能称出多少种重量。

#include<iostream>
using namespace std;

int dp[105][100005];
int a[19];
int main()
{
	
	int n,sum=0;cin >> n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	dp[0][0] = 1;
	for (int i = 1;i <= n;i++)
	{
		
		for (int j = 0;j <= sum;j++)
		{
			
			if (dp[i-1][j])
			{
				
				dp[i][j] = 1;
				dp[i][j + a[i]] = 1;
				dp[i][abs(j - a[i])] = 1;
			}
		}
	}
	int ans = 0;
	for (int i = 1;i <= sum;i++)if (dp[n][i]) ans++;
	cout << ans;
}