Perfect Squares (水题)

Perfect Squares


Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.

A number x is said to be a perfect square if there exists an integer y such that x = y2.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.

It is guaranteed that at least one element of the array is not a perfect square.

Output

Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.

Example
Input
2
4 2
Output
2
Input
8
1 2 4 8 16 32 64 576
Output
32
Note

In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

唯一需要注意的是给定的数有可能是负数,负数一定不是

coed:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
using namespace std;
int main(){
    int n;
    int a[1010];
    scanf("%d",&n);
    int i;
    for(i = 0; i < n; i++){
        scanf("%d",&a[i]);
    }
    int ans = -INF;
    for(i = 0; i < n; i++){
        int tmp = (int)sqrt((double)a[i]);
        if(a[i]<0||tmp*tmp!=a[i]){
            ans = max(ans,a[i]);
        }
    }
    printf("%d\n",ans);
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值