# Find ‘N’ number of solutions with the given inequality equations

• Last Updated : 21 Aug, 2022

Find the value of a1, a2, a3, ….an such that the following two conditions are satisfied.  Print the value of a1, a2, …, an and “No solution” otherwise.

Note: There maybe a several solutions, print any of them.

Examples:

Input: n = 5, x = 15, y = 15
Output:
11
1
1
1
1
Input: n = 4, x = 324, y = 77
Output:
74
1
1
1

Approach: Below is the step by step algorithm to solve this problem:

1. Initialize the number of elements and the value of x and y.
2. There is no solution of a1…a2 if y is less than n or if x is very larger than n.
3. Print first solution as y – n + 1 and 1 as the solution of rest of the elements.

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach #include  using namespace std; #define ll long long   // Function to calculate all the solutions void findsolution(ll n, ll x, ll y) {     // there is no solutions     if ((y - n + 1) * (y - n + 1) + n - 1 < x || y < n) {         cout << "No solution";         return;     }       // print first element as y-n+1     cout << y - n + 1;       // print rest n-1 elements as 1     while (n-- > 1)         cout << endl              << 1; }   // Driver code int main() {     // initialize the number of elements     // and the value of x an y     ll n, x, y;     n = 5, x = 15, y = 15;       findsolution(n, x, y);       return 0; }

## Java

 // java implementation of above approach import java.io.*;   class GFG {      // Function to calculate all the solutions static void findsolution(long n, long x, long y) {     // there is no solutions     if ((y - n + 1) * (y - n + 1) + n - 1 < x || y < n) {         System.out.println( "No solution");         return;     }       // print first element as y-n+1     System.out.println( y - n + 1);       // print rest n-1 elements as 1     while (n-- > 1)             System.out.println( "1"); }   // Driver code       public static void main (String[] args) {             // initialize the number of elements     // and the value of x an y     long n, x, y;     n = 5; x = 15; y = 15;       findsolution(n, x, y);     } } // This code is contributed  // by ajit

## Python3

 # Python3 implementation of above approach   # Function to calculate all the solutions def findsolution(n, x, y):       # there is no solutions     if ((y - n + 1) * (y - n + 1) +               n - 1 < x or y < n):         print("No solution");         return;       # print first element as y-n+1     print(y - n + 1);       # print rest n-1 elements as 1     while (n > 1):         print(1);         n -= 1;   # Driver code   # initialize the number of elements # and the value of x an y n = 5;  x = 15;  y = 15;   findsolution(n, x, y);   # This code is contributed by mits

## C#

 // C# implementation of above approach using System;   class GFG {       // Function to calculate all the solutions  static void findsolution(long n,                           long x, long y)  {      // there is no solutions      if ((y - n + 1) * (y - n + 1) +           n - 1 < x || y < n)      {          Console.WriteLine( "No solution");          return;      }        // print first element as y-n+1      Console.WriteLine( y - n + 1);        // print rest n-1 elements as 1      while (n-- > 1)          Console.WriteLine( "1");  }    // Driver code  static public void Main () {     // initialize the number of elements      // and the value of x an y      long n, x, y;      n = 5; x = 15; y = 15;            findsolution(n, x, y);  }  }    // This code is contributed  // by ajit

## PHP

  1)     echo "\n" . 1; }   // Driver code   // initialize the number of elements // and the value of x an y $n = 5; $x = 15; $y = 15; findsolution($n, $x, $y);   // This code is contributed  // by Akanksha Rai(Abby_akku)

## Javascript

 

Output:

11
1
1
1
1

Time Complexity: O(n)

Auxiliary Space: O(1)

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