找到满足给定方程的 n 个正整数

给定三个整数N 、 X和Y 。任务是找到满足给定方程的N 个正整数。

1. a ₁ ² + a ₂ ² + …. + a _n ² ≥ X
2. a ₁ + a ₂ + …. + a _n ≤ Y

如果不可能有这样的整数序列，则打印-1 。
例子：

Input: N = 3, X = 254, Y = 18 
Output: 1 1 16 
1² + 1² + 16² = 1 + 1 + 256 = 258 which is ≥ X 
1 + 1 + 16 = 18 which is ≤ Y
Input: N = 2, X = 3, Y = 2 
Output: -1 
No such sequence exists. 
 

方法：很容易看出，为了使平方和最大化，应该使除第一个数字以外的所有数字都等于1，并最大化第一个数字。记住这一点，我们只需要检查 y 的给定值是否足够大以满足所有 n 数都是正数的限制。如果 y 不是太小，那么我们只需要确保X ≤ 1 + 1 + … + (y – (n – 1)) ² 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find n positive integers
// that satisfy the given conditions
void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    vector ans;
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.push_back(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0) {
        cout << "-1";
        return;
    }
 
    // Place Nth integer
    ans.push_back(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += ans[i] * ans[i];
 
    // If it is less than x
    if (store < x) {
        cout << "-1";
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        cout << ans[i] << " ";
}
 
// Driver code
int main()
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Function to find n positive integers
// that satisfy the given conditions
static void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    ArrayList ans = new ArrayList();
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.add(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0)
    {
        System.out.print("-1");
        return;
    }
 
    // Place Nth integer
    ans.add(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += ans.get(i) * ans.get(i);
 
    // If it is less than x
    if (store < x)
    {
        System.out.print("-1");
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        System.out.print(ans.get(i)+" ");
}
 
// Driver code
public static void main (String[] args)
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
}
}
 
// This code is contributed by mits

Python3

# Python3 implementation of the approach
 
# Function to find n positive integers
# that satisfy the given conditions
def findIntegers(n, x, y):
 
    # To store n positive integers
    ans = []
 
    # Place N - 1 one's
    for i in range(n - 1):
        ans.append(1)
 
    # If can not place (y - (n - 1))
    # as the Nth integer
    if (y - (n - 1) <= 0):
        print("-1", end = "")
        return
 
    # Place Nth integer
    ans.append(y - (n - 1))
 
    # To store the sum of
    # squares of N integers
    store = 0
 
    for i in range(n):
        store += ans[i] * ans[i]
 
    # If it is less than x
    if (store < x):
        print("-1", end = "")
        return;
 
    # Print the required integers
    for i in range(n):
        print(ans[i], end = " ")
 
# Driver code
n, x, y = 3, 254, 18
findIntegers(n, x, y)
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
using System.Collections;
 
class GFG
{
     
// Function to find n positive integers
// that satisfy the given conditions
static void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    ArrayList ans = new ArrayList();
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.Add(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0)
    {
        Console.Write("-1");
        return;
    }
 
    // Place Nth integer
    ans.Add(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += (int)ans[i] *(int)ans[i];
 
    // If it is less than x
    if (store < x)
    {
        Console.Write("-1");
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        Console.Write((int)ans[i]+" ");
}
 
// Driver code
static void Main()
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
}
}
 
// This code is contributed by mits

Javascript

 
// JavaScript implementation of the approach
 
// Function to find n positive letegers
// that satisfy the given conditions
function findIntegers(n, x, y)
{
   
    // To store n positive letegers
    let ans = [];
   
    // Place N - 1 one's
    for (let i = 0; i < n - 1; i++)
        ans.push(1);
   
    // If can not place (y - (n - 1))
    // as the Nth leteger
    if (y - (n - 1) <= 0)
    {
        document.write("-1");
        return;
    }
   
    // Place Nth leteger
    ans.push(y - (n - 1));
   
    // To store the sum of
    // squares of N letegers
    let store = 0;
    for (let i = 0; i < n; i++)
        store += ans[i] * ans[i];
   
    // If it is less than x
    if (store < x)
    {
        document.write("-1");
        return;
    }
   
    // Print the required letegers
    for (let i = 0; i < n; i++)
        document.write(ans[(i)]+" ");
}     
     
// Driver Code
 
     let n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
           

输出：

1 1 16