​

一、朴素解法

#include <bits/stdc++.h>

using namespace std;
const int N = 10010;
typedef long long LL;
struct Person {
    int first, second;
} p[N];
int n;

bool cmp(const Person &a, const Person &b) {
    return a.first * a.second < b.second * b.first;
}

int main() {
    cin >> n >> p[0].first >> p[0].second; //n:表示大臣的人数;a,b:分别表示国王左手和右手上的整数
    for (int i = 1; i <= n; i++)cin >> p[i].first >> p[i].second;
    sort(p + 1, p + 1 + n, cmp);

    LL sum = 1;
    LL MAX = 0;
    for (int i = 0; i <= n; i++) {
        MAX = max(MAX, sum / p[i].second);
        sum *= p[i].first;
    }
    cout << MAX << endl;
    return 0;
}

二、高精度题解

#include<bits/stdc++.h>

using namespace std;

const int N = 1010;
int n;
struct Person {
    int left, right;
} person[N];

bool cmp(const Person &a, const Person &b) {
    return a.left * a.right < b.left * b.right;
}

//高精度乘法
vector<int> mul(vector<int> &A, int b) {
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || t; i++) {
        if (i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

//高精度除法
vector<int> div(vector<int> &A, int b, int &r) {
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i--) {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

//获取两个vector<int>中较大的那个
vector<int> max_vec(vector<int> a, vector<int> b) {
    if (a.size() > b.size()) return a;
    if (a.size() < b.size()) return b;
    if (vector<int>(a.rbegin(), a.rend()) > (vector<int>(b.rbegin(), b.rend()))) return a;
    return b;
}

int main() {
    cin >> n;
    //输入
    for (int i = 0; i <= n; i++) cin >> person[i].left >> person[i].right;
    //排序
    sort(person + 1, person + n + 1, cmp);

    //队伍中放入国王
    vector<int> sum(1, person[0].left);

    //结果
    vector<int> res(1, 0);
    int r;
    for (int i = 1; i <= n; i++) {
        res = max_vec(res, div(sum, person[i].right, r));
        sum = mul(sum, person[i].left);
    }
    //输出
    for (int i = res.size() - 1; i >= 0; i--) cout << res[i];
}