给定整数n,求离根号n最近的整数。

def func(n):     p = 0.618     i = 0     j = n     while i<=j:         mid = int(p*i + (1-p)*j)         if abs(mid*mid-n)<=abs((mid-1)*(mid-1)-n) and abs(mid*mid-n)<=abs((mid+1)*(mid+1)-n):             return mid         elif abs((mid-1)*(mid-1)-n)<=abs(mid*mid-n) and abs(mid*mid-n)<=abs((mid+1)*(mid+1)-n):             j = mid - 1         else:             i = mid + 1

一次找小于sqrt(n)中最大的，一次找大于之中最小的，再比较这样？

Sqrt(n)四舍五入不行吗？

牛顿割线法

def find_nearest_n(number):     left = 0     right = number     while right-left >= 0.01:         if ((left+right)/2)**2 == number:             return (left+right)/2         elif ((left+right)/2)**2 > number:             right = (left+right)/2         else:             left = (left+right)/2     if int(right) > int(left):         return int(right)     elif left - int(left) < 0.5:         return int(left)     else:         return int(left)+1 阈值可以再调大一点 应该有还有更好的方法

import numpy as np def bin_search_sqrt(num): left, right = 0, num while right - left > 1: mid = (left + right) // 2 if mid ** 2 > num: right = mid else: left = mid return right if right**2 - num < num - left**2 else left for i in range(15, 37): print(i,bin_search_sqrt(i),np.sqrt(i))输出： 16 4 4.0 17 4 4.12310562562 18 4 4.24264068712 19 4 4.35889894354 20 4 4.472135955 21 5 4.58257569496 22 5 4.69041575982 23 5 4.79583152331 24 5 4.89897948557 25 5 5.0 26 5 5.09901951359 27 5 5.19615242271 28 5 5.29150262213 29 5 5.38516480713 30 5 5.47722557505 31 6 5.56776436283 32 6 5.65685424949 33 6 5.74456264654 34 6 5.83095189485 35 6 5.9160797831 36 6 6.0看上去好像是可以的？

  private static int core(int n){     int small=0,big=n/2+1,mid=0;     while(small<big){         mid = small+(big-small)/2;         if(mid*mid>n)  big = mid-1;         else if(mid*mid<n)  small=mid+1;         else   return mid;     }     int s=0,e=0;     if(mid*mid>n){         s = mid-1;e = mid;     }else{         s = mid;e = mid+1;     }     double m = (s+e)/2.0;     if(m*m<n) return e;     else return s; }