题解 | #最小邮票数#

用dfs比较简单
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

//利用深度优先遍历

int m, n;
int ans = 100;


void DFS(int pos,int stamps[], int current,int c ) {  // current表当前的价值，count表当前的数量
	if (pos == n-1) return;
	
	for (int i = pos+1; i < n; i++)   //遍历其邻居 
	{
		if (current+stamps[i] == m ) 
			ans = min(ans, c+1);

		if (current + stamps[i] < m) 
		{  //如果当前面额大于该邮票
			DFS(i, stamps, current + stamps[i], c + 1);
		}


	}

}


int main() {
	
	while (cin >> m >>n) 
	{
		int stamps[25];
		for (int i = 0; i < n; i++) {
			cin >> stamps[i];
		}
		DFS(0, stamps, 0, 0);
        if(ans==100) cout << "0"<<endl;
        else
		    cout << ans << endl;
	}

}