这个代码怎么写呀，一直显示超时

这个long long究竟放在哪里，为什么我改了x的类型还是超时

这个估计你不能循环从1开始进行判断，可以考虑用二分法折半查找。比如在1到x/2之间进行折半比较，判断平方数与n的关系，直到遇到相等数，或者比n大的和比n小的数之差是1则结束

#include <stdio.h>
#include <stdlib.h>
int main()
{
    long long x,l,h,left,right,mid;
    int  n,i;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%lld",&x);
        left = 1;
        right = x/2;
        h = right;
        l = left;
        while(left < right)
        {
            mid = (left + right)/2;
            if(mid * mid == x)
            {
                printf("%lld * %lld = %lld\n",mid,mid,mid*mid);
                break;
            }
            else if(mid * mid < x)
            {
                if(mid == h-1)
                {
                    printf("%lld * %lld = %lld\n",mid,mid,mid*mid);
                    break;
                }
                left = mid;
                l = left;
            }
            else
            {
                if(mid == l+1)
                {    
                    printf("%lld * %lld = %lld\n",l,l,l*l);
                    break;
                }
                right = mid;
                h = right;
            }
        }
    }
}

可以使用迭代算法计算平方根（参考https://en.wikipedia.org/wiki/Square_root#Computation
迭代公式是

#include <iostream>

unsigned long long my_sqrt(unsigned long long x) {
  if (x == 0)
    return 0;
  if (x == 1)
    return 1;
  unsigned long long x_n = x / 2;
  while (1) {
    unsigned long long x_n1 = (x_n + x / x_n) / 2;
    if (x_n1 >= x_n)
      break;
    x_n = x_n1;
  }
  return x_n;
}

int main() {
  unsigned long long x;
  while (std::cin >> x)
    std::cout << my_sqrt(x) << '\n';
  return 0;
}