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

• Last Updated : 15 Apr, 2021

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

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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 solutionsvoid 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 codeint 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 approachimport java.io.*; class GFG {    // Function to calculate all the solutionsstatic 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 solutionsdef 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 yn = 5;x = 15;y = 15; findsolution(n, x, y); # This code is contributed by mits

## C#

 // C# implementation of above approachusing System; class GFG{     // Function to calculate all the solutionsstatic 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 codestatic 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

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