查找满足给定方程的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 +…。 +一个_N ^2≥X
2. 一个₁ +一个₂ +…。 +_正≤ÿ

如果没有这样的整数序列，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，为了最大化平方和，应该使除第一个数字以外的所有数字都等于1，并使第一个数字最大化。牢记这一点，我们只需要检查y的给定值是否足够大以满足所有n个数均为正的限制。如果y不太小，那么我们需要确保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

输出：

1 1 16

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