题解 | #最小邮票数#

#include <bits/stdc++.h>
using namespace std;

const int N = 20;
int v[N];
int m,n,minK; //总值，邮票数量，最小数量

//当前的邮票编号，当前总价值，当前邮票数量
void dfs(int index,int sum,int k){
    if(sum == m){
        if(k < minK){
            minK = k;
            return;
        }
    }

    if(index >= n or sum > m) return ;

    //穷举，对任意一张邮票，走遍选择它与不选择它的两种可能。
    dfs(index + 1, sum + v[index] , k + 1);
    dfs(index + 1, sum , k );
}

int main(){
    while(cin >> m >> n){
        //初始化最小数量为邮票总数
        minK = n;

        for(int i = 0 ; i < n ; i ++) cin >> v[i];
        
        dfs(0,0,0);

        if(minK == n) cout << 0 << endl;
        else cout << minK << endl;
    }
}



#include <bits/stdc++.h>
using namespace std;
const int MAXM = 101;
const int MAXN = 21;
const int Max = 99999;
int dp[MAXN][MAXM];  

int main() {
    for (int i = 0; i < MAXN; i++) {
        for (int j = 0; j < MAXM; j++) {
            dp[i][j] = Max;
        }
    }
    for (int i = 0; i < MAXN; i++) dp[i][0] = 0;

    int M, N;
    cin >> M >> N;
    int v[N];
    for (int i = 0; i < N; i++) cin >> v[i];  

    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++) {
            if (v[i - 1] <= j) {
                dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - v[i - 1]] + 1);
            } else {
                dp[i][j] = dp[i - 1][j];
            }
        }
    }
    if (dp[N][M] != Max) cout << dp[N][M] << endl;
    else cout << 0 << endl;
    return 0;
}