最少要加或减的数，使其成为一个完美的正方形
给定一个数 N ，求 N 需要加减的最小数，使其成为完美正方形。如果要加数字，用+号打印，否则如果要减数字，用–号打印。
示例:
输入: N = 14 输出:2 14 = 9 14 = 16 后最近的完美方块因此，需要将 2 加到 14 上才能得到最近的完美方块。
输入: N = 18 输出:-2 18 = 16 18 = 25 后最近的完美正方形因此，需要从 18 中减去 2 才能得到最近的完美正方形。
接近:
- 拿到号码。
- 求数字的平方根，将结果转换为整数。
- 将双精度值转换为整数后，将包含 N 以上完美平方的根，即楼(平方根(N)) 。
- 然后求这个数的平方，这将是 n 之前的完美平方。
- 求 N 后完美平方的根，即天花板(平方根(N)) 。
- 然后求这个数的平方，这将是 n 之后的完美平方。
- 检查楼层值的平方是最接近 N 还是天花板值。
- 如果楼层值的平方最接近 N，用-号打印差值。否则，用+号打印天花板值的平方和 N 之间的差值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the Least number
int nearest(int n)
{
// Get the perfect square
// before and after N
int prevSquare = sqrt(n);
int nextSquare = prevSquare + 1;
prevSquare = prevSquare * prevSquare;
nextSquare = nextSquare * nextSquare;
// Check which is nearest to N
int ans
= (n - prevSquare) < (nextSquare - n)
? (prevSquare - n)
: (nextSquare - n);
// return the result
return ans;
}
// Driver code
int main()
{
int n = 14;
cout << nearest(n) << endl;
n = 16;
cout << nearest(n) << endl;
n = 18;
cout << nearest(n) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the Least number
static int nearest(int n)
{
// Get the perfect square
// before and after N
int prevSquare = (int)Math.sqrt(n);
int nextSquare = prevSquare + 1;
prevSquare = prevSquare * prevSquare;
nextSquare = nextSquare * nextSquare;
// Check which is nearest to N
int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
// return the result
return ans;
}
// Driver code
public static void main (String[] args)
{
int n = 14;
System.out.println(nearest(n));
n = 16;
System.out.println(nearest(n)) ;
n = 18;
System.out.println(nearest(n)) ;
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
from math import sqrt
# Function to return the Least number
def nearest(n) :
# Get the perfect square
# before and after N
prevSquare = int(sqrt(n));
nextSquare = prevSquare + 1;
prevSquare = prevSquare * prevSquare;
nextSquare = nextSquare * nextSquare;
# Check which is nearest to N
ans = (prevSquare - n) if (n - prevSquare) < (nextSquare - n) else (nextSquare - n);
# return the result
return ans;
# Driver code
if __name__ == "__main__" :
n = 14;
print(nearest(n)) ;
n = 16;
print(nearest(n));
n = 18;
print(nearest(n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG {
// Function to return the Least number
static int nearest(int n)
{
// Get the perfect square
// before and after N
int prevSquare = (int)Math.Sqrt(n);
int nextSquare = prevSquare + 1;
prevSquare = prevSquare * prevSquare;
nextSquare = nextSquare * nextSquare;
// Check which is nearest to N
int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
// return the result
return ans;
}
// Driver code
public static void Main (string[] args)
{
int n = 14;
Console.WriteLine(nearest(n));
n = 16;
Console.WriteLine(nearest(n)) ;
n = 18;
Console.WriteLine(nearest(n)) ;
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to return the Least number
function nearest( n)
{
// Get the perfect square
// before and after N
var prevSquare = parseInt(Math.sqrt(n));
var nextSquare = prevSquare + 1;
prevSquare = prevSquare * prevSquare;
nextSquare = nextSquare * nextSquare;
// Check which is nearest to N
if((n - prevSquare) < (nextSquare - n))
{
ans = parseInt((prevSquare - n));
}
else
ans = parseInt((nextSquare - n));
// return the result
return ans;
}
var n = 14;
document.write( nearest(n) + "<br>");
n = 16;
document.write( nearest(n) + "<br>");
n = 18;
document.write( nearest(n) + "<br>");
// This code is contributed by SoumikMondal
</script>
Output:
2
0
-2
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