题目链接

题目链接

solution

通过拆解公式，可以得出当a[i] < x时候，进行公式变换后也还是原来的数值，所以我们只用处理大于等于的情况。首先想到的方法是搜索。

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<bitset>
#include<cstdio>
#include<cstring>
#include<fstream>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<unordered_map>

using namespace std;
#define x first
#define y second
#define endl '\n'
#define IOS                       \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f, mod = 1000000007;
const int N = 2e5 + 10;
ll ans = 1e18, x, n, a[30];
vector<int> v;
void dfs(ll last, ll x)
{
    if (last >= v.size() || (last < v.size() && x > v[v.size() - 1])) {
        ans = min(x, ans);
        return;
    }
  
    for (int i = last; i < v.size(); i++) {
        if (v[i] >= x) {
            dfs(i + 1, x + (v[i] / x) * (v[i] + v[i] % x));
        }
    }
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    cin >> x;
    for (int i = 1; i <= n; i++) {
        if (a[i] >= x) {
            v.emplace_back(a[i]);
        }
    }
    sort(v.begin(), v.end());
    dfs(0, x);
    cout << ans << endl;
}
int main(){
    // IOS;
    int t = 1;
    // cin >> t;
    while(t--){
        solve();
    }
    return 0;
}