题目链接
Let’s call an array of non-negative integers a1,a2,…,an a k-extension for some non-negative integer k if for all possible pairs of indices 1≤i,j≤n the inequality k⋅|i−j|≤min(ai,aj) is satisfied. The expansion coefficient of the array a is the maximal integer k such that the array a is a k-extension. Any array is a 0-expansion, so the expansion coefficient always exists.
You are given an array of non-negative integers a1,a2,…,an. Find its expansion coefficient.
Input
The first line contains one positive integer n — the number of elements in the array a (2≤n≤300000). The next line contains n non-negative integers a1,a2,…,an, separated by spaces (0≤ai≤109).
Output
Print one non-negative integer — expansion coefficient of the array a1,a2,…,an.
input
4
6 4 5 5
output
1
input
3
0 1 2
output
0
input
4
821 500 479 717
output
239
Note
In the first test, the expansion coefficient of the array [6,4,5,5] is equal to 1 because |i−j|≤min(ai,aj), because all elements of the array satisfy ai≥3. On the other hand, this array isn’t a 2-extension, because 6=2⋅|1−4|≤min(a1,a4)=5 is false.
In the second test, the expansion coefficient of the array [0,1,2] is equal to 0 because this array is not a 1-extension, but it is 0-extension.
题意：给你n个数，让你找到最大的一个数K，使得K*|i - j| <= min(ai,aj) 对于任意的1 <= i,j <= n皆成立（i,j）是下标
题解：对于给定的每一个数a把它当作最小值，判断距离左右端点的最大值d，然后计算ans = a / d，遍历一遍找到最小值集合中的真正的最小值
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m,d,ans = INT_MAX;
scanf("%d",&n);
for(int i = 1;i <= n;i++){
d = max(i - 1,n - i);
scanf("%d",&m);
ans = min(ans,m / d);
}
printf("%d\n",ans);
return 0;
}