题意翻译
给定一组有n个整数的数组a1,a2,…,an.找出这组数中的最大非完全平方数。 完全平方数是指有这样的一个数x,存在整数y,使得x=y^2y2 .
输入格式
第一行输入一个单独的整数n(1<=n<=1000 )，n为该组数的数量。 接下来有n个整数，a1,a2,an(-10^6<=a<=10^6) 数据保证至少存在一个整数是非完全平方数。
题目描述
Given an array a_{1},a_{2},...,a_{n}a1?,a2?,...,an? of nn integers, find the largest number in the array that is not a perfect square.
A number xx is said to be a perfect square if there exists an integer yy such that x=y^{2}x=y2 .
输入输出格式
输入格式：
The first line contains a single integer nn ( 1<=n<=10001<=n<=1000 ) — the number of elements in the array.
The second line contains nn integers a_{1},a_{2},...,a_{n}a1?,a2?,...,an? ( -10^{6}<=a_{i}<=10^{6}−106<=ai?<=106 ) — the elements of the array.
It is guaranteed that at least one element of the array is not a perfect square.
输出格式：
Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.
输入输出样例
说明
In the first sample case, 44 is a perfect square, so the largest number in the array that is not a perfect square is 22 .
#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 1001 using namespace std; int n; int num[MAXN]; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&num[i]); sort(num+1,num+1+n); for(int i=n;i>=1;i--){ int k=sqrt(num[i]); if(k*k!=num[i]){ cout<<num[i]; return 0; } } }
