반복해서 쓸 수 있는 이동량

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int solution(vector<int> steps, int target) {
    const int INF = 1000000;
    vector<int> dp(target + 1, INF);
    dp[0] = 0;
    for (int step : steps) {
        for (int x = target; x >= step; --x) {
            if (dp[x - step] != INF) dp[x] = min(dp[x], dp[x - step] + 1);
        }
    }
    return dp[target] == INF ? -1 : dp[target];
}
int main() {
    int target, n; cin >> target >> n;
    vector<int> steps(n);
    for (int i = 0; i < n; ++i) cin >> steps[i];
    cout << solution(steps, target) << '\n';
}