为什么我的 C++ 最大化成对乘积代码不起作用？

Question

So the problem asked me to find the greatest product from a given sequence of non-negative integers. What I did was I tried to find the greatest two integers from the sequence (for which I used a vector, by taking an input of n numbers) and multiplied them, since there are no negative integers. I used the long long type as well, since just an int type would not be enough for huge numbers. I kept getting a random huge number as the output whenever I tried to run the program:

#include <iostream>
#include <vector>
using namespace std;

long long max_prod(const vector<int>& numbers) {
    int max1 = -1;
    int max2 = -1;
    int n = numbers.size();
    for (int i = 0; i<n; i++){
        if (numbers[i] > numbers[max1]) 
            max1 = i;
    }
    for (int j = 0; j < n; j++) {
        if (numbers[j] > numbers[max2] && j!=max1)
            max2 = j;
    }
return ((long long)(numbers[max1])) * ((long long)(numbers[max2]));
}

int main(){
    int n;
    cin >> n;
    vector<int> numbers(n);
    for (int i = 0; i<n; i++){
        cin >> numbers[i];
    }
    long long result = max_prod(numbers);
    cout << result << "\n";
    return 0;
}

the last line is the output given by the program

Answer 1

You haver undefined behavior right here

long long max_prod(const vector<int>& numbers) {
    int max1 = -1; <<<<<====
    int max2 = -1;
    int n = numbers.size();
    for (int i = 0; i < n; i++) {
        if (numbers[i] > numbers[max1]) <<<<<==
            max1 = i;
    }
    for (int j = 0; j < n; j++) {
        if (numbers[j] > numbers[max2] && j != max1)
            max2 = j;
    }
    return ((long long)(numbers[max1])) * ((long long)(numbers[max2]));
}

You try to access numbers[-1] (twice once in the j loop and once in the i loop).

Set maxi and maxj to 0