解题思路：先循环找一次最大的a_ib_i，然后循环把所有等于该最大值的菜品标记售罄。然后在剩下的菜里找最大的a_ib_i，如果等大，找最小的|a_i-b_i|。只要是顺着遍历的，找到的一定是编号最小的。

注意10^9是会爆int的，所以建议使用long long；另外，本题如果用冒泡排序等一般的排序来做，会超时。(快速排序等\Omicron(n\log n)的排序不会超时，有兴趣可以自行了解)

参考代码：

C语言:

#include <stdio.h>
#include <math.h>
long long n, a[100010], b[100010], max_ab = -1, min_d = 1e10, k, v[100010];
int main()
{
    scanf("%lld", &n);
    for (int i = 1; i <= n; ++i) //为让下标与编号对应，从下标1开始使用数组
    {
        scanf("%lld", &a[i]);
    }
    for (int i = 1; i <= n; ++i)
    {
        scanf("%lld", &b[i]);
    }
    for (int i = 1; i <= n; ++i)
    {
        if (a[i] * b[i] > max_ab)
        {
            max_ab = a[i] * b[i];
        }
    }
    for (int i = 1; i <= n; ++i)
    {
        if (a[i] * b[i] == max_ab)
        {
            v[i] = 1; //表示售罄
        }
    }
    max_ab = -1; //重新初始化
    for (int i = 1; i <= n; ++i)
    {
        if (v[i] == 0 && (a[i] * b[i] > max_ab || a[i] * b[i] == max_ab && fabs(a[i] - b[i]) < min_d))
        {
            max_ab = a[i] * b[i];      //除去售罄菜外当前最美味的值
            min_d = fabs(a[i] - b[i]); //在最美味的值里，最小|a_i-b_i|的差值
            k = i;
        }
    }
    if (k == 0) //全局变量初始值为0
    {
        printf("sold out");
    }
    else
    {
        printf("%lld %lld", k, a[k] * b[k]);
    }
    return 0;
}

Python:

n = int(input())
a = [int(i) for i in input().strip().split(' ')]
b = [int(i) for i in input().strip().split(' ')]
max_ab = -1
min_d = 1e10
ans = -1
for i in range(n):
    if a[i]*b[i] > max_ab:
        max_ab = a[i]*b[i]
v = [a[i]*b[i] == max_ab for i in range(n)]
max_ab = -1
for i in range(n):
    if not v[i] and (a[i]*b[i] > max_ab or a[i]*b[i] == max_ab and abs(a[i]-b[i]) < min_d):
        ans = i
        max_ab = a[i]*b[i]
        min_d = abs(a[i]-b[i])
if ans == -1:
    print('sold out')
else:
    print('%d %d'%(ans+1, a[ans]*b[ans]))