无找零购物(No Change G) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int k, n;
    if (!(cin >> k >> n)) return 0;
    vector<long long> coin(k);
    for (int i = 0; i < k; ++i) cin >> coin[i];
    vector<long long> pref(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        long long c;
        cin >> c;
        pref[i] = pref[i - 1] + c;
    }
    int tot = 1 << k;
    vector<int> dp(tot, 0);
    long long ans = -1;
    for (int mask = 0; mask < tot; ++mask) {
        int cur = dp[mask];
        if (cur == n) {
            long long rem = 0;
            for (int i = 0; i < k; ++i) {
                if (!((mask >> i) & 1)) rem += coin[i];
            }
            ans = max(ans, rem);
        }
        for (int i = 0; i < k; ++i) {
            if (!((mask >> i) & 1)) {
                long long target = pref[cur] + coin[i];
                int nxt = (int)(upper_bound(pref.begin(), pref.end(), target) - pref.begin()) - 1;
                int nmask = mask | (1 << i);
                if (nxt > dp[nmask]) dp[nmask] = nxt;
            }
        }
    }
    cout << ans << "\n";
    return 0;
}