找到最小的完美平方数 A，这样 N + A 也是一个完美平方数
原文:https://www . geesforgeks . org/find-minist-perfect-square-number-a-so-n-a-也是-perfect-square-number/
给定一个正数 N 。任务是找出最小的完美平方数 A，使得 N + A 也是一个完美平方数或返回-1。 例:
Input: N = 3
Output: 1
Explanation:
As 1 + 3 = 4 = 22
Input: N=1
Output: -1
天真方法: 从{1，2，3，4，5…}遍历 M，检查(N + M * M)是否是一个完美的平方数。 高效方法:
- 在观察时，我们有一个方程，如:
- N+(X * X) = (M * M) where n is given, m and x are unknown.
- We can rearrange it to get:
- n =(M * M)–(X * X)
- n =(M+X)*(M–X)
-
现在我们可以看到，为了得到 N，我们需要找到 N 的因子，N 的因子可以在 O(N)时间内得到。但是可以通过这个方法优化到 O(N^1/2)。
-
设 N 的因子为 a 和 b = (N / a)。所以，从上面的方程 a =(M–X)和 b = (M + X)，求解后我们可以得到X =(b–a)/2的值。
以下是上述方法的实现:
C++
// C++ code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
#include<bits/stdc++.h>
using namespace std;
long SmallestPerfectSquare(long N){
// X is the smallest perfect
// square number
long X = (long)1e9;
long ans;
// Loop from 1 to square root of N
for(int i = 1; i < sqrt(N); i++)
{
// Condition to check whether i
// is factor of N or not
if (N % i == 0)
{
long a = i;
long b = N / i;
// Condition to check whether
// factors satisfies the
// equation or not
if((b - a != 0) && ((b - a) % 2 == 0))
{
// Stores minimum value
X = min(X, (b - a) / 2);
}
}
}
// Return if X * X if X is not equal
// to 1e9 else return -1
if (X != 1e9)
ans = X * X;
else
ans = -1;
return ans;
}
// Driver code
int main()
{
long N = 3;
cout << SmallestPerfectSquare(N);
return 0;
}
// This code is contributed by AnkitRai01
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
public class GFG {
static long SmallestPerfectSquare(long N)
{
// X is the smallest perfect
// square number
long X = (long)1e9;
long ans;
// Loop from 1 to square root of N
for(int i = 1; i < Math.sqrt(N); i++){
// Condition to check whether i
// is factor of N or not
if (N % i == 0){
long a = i ;
long b = N / i;
// Condition to check whether
// factors satisfies the
// equation or not
if ((b - a != 0) && ((b - a) % 2 == 0)){
// Stores minimum value
X = Math.min(X, (b - a) / 2) ;
}
}
}
// Return if X * X if X is not equal
// to 1e9 else return -1
if (X != 1e9)
ans = X * X;
else
ans = -1;
return ans;
}
// Driver code
public static void main (String[] args){
long N = 3;
System.out.println(SmallestPerfectSquare(N)) ;
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 code to find out the smallest
# perfect square X which when added to N
# yields another perfect square number.
import math
def SmallestPerfectSquare(N):
# X is the smallest perfect
# square number
X = 1e9
# Loop from 1 to square root of N
for i in range(1, int(math.sqrt(N)) + 1):
# Condition to check whether i
# is factor of N or not
if N % i == 0:
a = i
b = N // i
# Condition to check whether
# factors satisfies the
# equation or not
if b - a != 0 and (b - a) % 2 == 0:
# Stores minimum value
X = min(X, (b - a) // 2)
# Return if X * X if X is not equal
# to 1e9 else return -1
return(X * X if X != 1e9 else -1)
# Driver code
if __name__ == "__main__" :
N = 3
print(SmallestPerfectSquare(N))
C
// C# code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
using System;
class GFG {
static long SmallestPerfectSquare(long N)
{
// X is the smallest perfect
// square number
long X = (long)1e9;
long ans;
// Loop from 1 to square root of N
for(int i = 1; i < Math.Sqrt(N); i++){
// Condition to check whether i
// is factor of N or not
if (N % i == 0)
{
long a = i;
long b = N / i;
// Condition to check whether
// factors satisfies the
// equation or not
if ((b - a != 0) && ((b - a) % 2 == 0))
{
// Stores minimum value
X = Math.Min(X, (b - a) / 2);
}
}
}
// Return if X*X if X is not equal
// to 1e9 else return -1
if (X != 1e9)
ans = X * X;
else
ans = -1;
return ans;
}
// Driver code
public static void Main (string[] args)
{
long N = 3;
Console.WriteLine(SmallestPerfectSquare(N));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// JavaScript code to find out the smallest
// perfect square X which when added to N
// yields another perfect square number.
function SmallestPerfectSquare(N){
// X is the smallest perfect
// square number
let X = 1e9;
let ans;
// Loop from 1 to square root of N
for(let i = 1; i < Math.sqrt(N); i++)
{
// Condition to check whether i
// is factor of N or not
if (N % i == 0)
{
let a = i;
let b = N / i;
// Condition to check whether
// factors satisfies the
// equation or not
if((b - a != 0) && ((b - a) % 2 == 0))
{
// Stores minimum value
X = Math.min(X, (b - a) / 2);
}
}
}
// Return if X * X if X is not equal
// to 1e9 else return -1
if (X != 1e9)
ans = X * X;
else
ans = -1;
return ans;
}
// Driver code
let N = 3;
document.write(SmallestPerfectSquare(N));
// This code is contributed by Surbhi Tyagi.
</script>
Output:
1
时间复杂度: sqrt(N)
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