AtCoder-abc231_e 题解

发表于2023-10-13更新于2023-10-13

题目传送门

一道贪心题……

感觉很裸啊，模拟赛时随便乱写了个暴力递归就能过。每次找最接近钱数 xx 的面额 numnum，如果比钱数少那么答案为剩下 x mod numx \bmod num 钱数的答案加上 x÷numx \div num。否则答案则为剩下 num−xnum-x 钱数的答案加上 11。

Code

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
ll n, m;
ll a[65];
ll ans(ll x) {
    if (x == 0) return 0;
    ll minn = 1e18, num;
    for (int i = 1; i <= n; i++) {
        if (llabs(x - a[i]) < minn) {
            minn = llabs(x - a[i]);
            num = a[i];
        }
    }
    if (num > x) return ans(num - x) + 1;
    return ans(x % num) + x / num;
}
int main() {
    ios :: sync_with_stdio(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    cout << ans(m);
    return 0;
}