题解 | #平方根#

方法一：

```// 牛顿送代法

#include<bits/stdc++.h>

using namespace std;

int n;
double f( int m )
{
	double last = m;
	double ans = m;
	ans = ( ans + m / ans ) / 2;
    
	while( abs( ans - last ) > 0.0001 )
    {
		last = ans;
		ans = ( ans + m / ans ) / 2;
	}
    
	return ans;
}

int main()
{
    cin >> n;
    cout << int(f(n)) << endl;
}

法二：

```// 二分法

#include<bits/stdc++.h>

using namespace std;

long long n, c, cnt, st;

int main()
{
    cin >> c;
    
    int l = 0, r = c;
    
    while( l < r )
	{
        int mid = l + r >> 1;
        
        if( mid * mid > c )
		{
			r = mid - 1;
		}
        else if( mid * mid < c )
		{
			l = mid + 1;
		}
        else
		{
            cout << int(mid);
            return 0;
        }
    }
    
    if( r > l )
	{
        if( r * l < c && ( r + 1 ) * ( r + 1 ) < c )
		{
			cout << int( r + 1 );
		}
        else if( r * r < c && ( r + 1 ) * ( r + 1 ) > c )
		{
			cout << int(r);
		}
        else if( r * r > c )
		{
			cout << int( r - 1 );
		}
        else
		{
			cout << int(r);
		}
    }
    else
	{
        if( l * l < c && ( l + 1 ) * ( l + 1 ) < c )
		{
			cout << int( l + 1 );
		}
        else if( l * l < c && ( l + 1 ) * ( l + 1 ) > c )
		{
			cout << int(l);
		}
        else if( l * l > c )
		{
			cout << int( l - 1 );
		}
        else
		{
			cout << int(l);
		}
    }
      
    return 0;
}

法三：

``` // 约翰 - 卡马克算法

#include<bits/stdc++.h>

using namespace std;

int n;
float f(float x)
{
    float xhalf = 0.5f * x;
    int i = *(int*) & x;
    i = 0x5f375a86 - ( i >> 1 );
    x = *(float*) & i;
    x = x * ( 1.5f - xhalf * x * x );
    x = x * ( 1.5f - xhalf * x * x );
    x = x * ( 1.5f - xhalf * x * x );

    return 1 / x;
}


int main()
{
    cin >> n;
    cout << int(f(n)) << endl;
    return 0;
}