// dp[i][j]表示只用前i种邮票能凑成总值M的最少邮票数,如果凑不成为INF // dp[0][j] = INF // dp[i][0] = 0 /* dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - p[i]] + 1) */ #include<bits/stdc++.h> using namespace std; const int INF = 1e8; int main() { int M, n; while (cin >> M >> n) { vector<int> p(n, 0); for (int i = 0;i < n;i++) cin >> p[i]; vector<int> dp(M + 1, INF); dp[0] = 0; for (int i = 0;i < n;i++) for (int j = M;j >= p[i];j--) dp[j] = min(dp[j], dp[j - p[i]] + 1); cout << (dp[M] == INF ? 0 : dp[M]) << endl; } return 0; }