题目描述

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样例

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
// #define int long long

using namespace std;

typedef pair<int, int > PII;
PII man[1010];
// A > B
bool cmp(vector<int> &A, vector<int> &B){
    if(A.size() != B.size()) return A.size() > B.size();
    for(int i = A.size() - 1; i >= 0; i--){
        if(A[i] != B[i]) return A[i] > B[i];
    }
    return false;
}

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

vector<int> div(vector<int> &A, int b){
    int t = 0;
    vector<int> C;
    for(int i = A.size()-1; i >= 0; i--){
        t = t*10 + A[i];
        C.push_back(t/b);
        t %= b;
    }

    reverse(C.begin(), C.end());
    while(C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

signed main(){
    int n;
    cin >> n;

    for(int i = 1; i <= n+1; i++){
        int l, r;
        cin >> l >> r;
        man[i] = {l*r, l};
    }
    vector<int> sum, res;
    sum.push_back(1);
    // int res = -1e9;
    sort(man+2, man+2+n);
    if(n == 1){
        cout << man[1].second / (man[2].first/man[2].second);
        return 0;
    }
    for(int i = 1; i <= n+1; i++){
        int l = man[i].second;
        int r = man[i].first / man[i].second;
        //res = max(res, sum/r);
        auto tmp = div(sum, r);
        if( cmp(tmp, res) )
        {
            res.clear();
            for(int i = 0; i < tmp.size(); i++){
                res.push_back(tmp[i]);
            }
        }
        sum = mul(sum, l);
        // cout << l << " " << r << "|" << sum << "|" << res << "\n"; 
    }
    for(int i = res.size() - 1; i >= 0; i--){
        cout << res[i];
    }
    // vector<int> tmp;
    // tmp.push_back(1);
    // auto C = div(tmp, 10);
    // for(int i = C.size() - 1; i >= 0 ; i--){
    //     cout << C[i];
    // }
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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