题解 | #最小邮票数#

题解都是一维的，来个比较容易理解的朴素的二维

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 105;

int n, m, v[N], f[N][N];

int main()
{
    cin >> m >> n;
    memset(f, 0x3f, sizeof(f));
    for (int i = 0; i <= n; ++i)
        f[i][0] = 0;
    for (int i = 1; i <= n; ++i)
        cin >> v[i];
    for (int i = 1; i <= n; ++i)
        for (int j = 0; j <= m; ++j)
            if (j >= v[i])
                f[i][j] = min(f[i-1][j], f[i-1][j-v[i]] + 1);
            else
                f[i][j] = f[i-1][j];
    if (f[n][m] == 0x3f3f3f3f)
        cout << 0 << endl;
    else
        cout << f[n][m] << endl;
    return 0;
}