ZJ c471

// Judge: https://zerojudge.tw/ShowProblem?problemid=c471

#include <bits/stdc++.h>
using namespace std;

// This problem is a little harder than previous problems.
// The idea is about greedy.
// Let us consider situation when there're only two stuff: stuff A and B
// If A stack on B, it costs f(A) * w(B), otherwise it costs f(B) * w(A).
// Formally, if (f(A) * w(B) < f(B) * w(A)), we should let A stack on B to minimize the cost, and vice versa.
//
// Now we can extend to N stuff and sort them, and than calculate the answer!

struct stuff{
    int w,f; //w => weight, f => times to take
};
stuff stuffs[100000];

// compare function, for sorting usage
bool cmp(stuff a, stuff b){
    return a.w*b.f < b.w*a.f;
}

int main(){
    // inputs
    int n;
    cin >> n;
    for(int i = 0; i < n; ++i){
        cin >> stuffs[i].w;
    }
    for(int i = 0; i < n; ++i){
        cin >> stuffs[i].f;
    }
    // sort method, see more details https://officeguide.cc/cpp-sort-function-tutorial-examples/
    sort(stuffs, stuffs + n, cmp);
    // declare the answer and the sum of weight
    // remember to use long long int to prevent from integer overflow
    long long int ans = 0, sum = 0;
    for(int i = 0; i < n; ++i){
        // calculation
        ans += sum * stuffs[i].f;
        sum += stuffs[i].w;
    }
    cout << ans << '\n';
    return 0;
}