這道題就是一道簡單的二維dp，轉移方程為dp[i][j] = min(dp[i-a[j]][j-1]+1, dp[i][j]);//i>a[j]時
                                             dp[i][j] = min(dp[i][j], dp[i][j-1]);

其中dp[i][j]表示用前j張郵票湊成數額為i的最小郵票數目，附上AC程式碼：

#include<bits/stdc++.h>
using namespace std; 
int dp[110][25];
int a[25];
int main()
{
	int m, n;
	while(~scanf("%d%d", &m, &n))
	{
		for(int i = 1; i <= n; i++)
		   scanf("%d", &a[i]);
		for(int i = 1; i <= m; i++)
		   for(int j = 1; j <= n; j++)
		      dp[i][j] = 1e9;
	    for(int i = 1; i <= n; i++)
		    dp[a[i]][i] = 1;
		
		for(int i = 1; i <= m; i++)
		{
			for(int j = 2; j <= n; j++)
			{
				//printf("aaa\n");
				if(i > a[j])
				   dp[i][j] = min(dp[i-a[j]][j-1]+1, dp[i][j]);
				dp[i][j] = min(dp[i][j], dp[i][j-1]);
				//printf("dp[%d][%d]:%d\n", i, j, dp[i][j]);
			}
		}
		if(dp[m][n] == 1e9)
		   printf("0\n");
		else 
		   printf("%d\n", dp[m][n]);
	}
	return 0;
}