Given 3 integers K, P and N. Where, K is the number of problems given to the person every day and P is the maximum number of problems he can solve in a day. Find the total number of problems not solved after the N-th day.
Examples:
Input : K = 2, P = 1, N = 3 Output : 3 On each day 1 problem is left so 3*1 = 3 problems left after Nth day. Input : K = 4, P = 1, N = 10 Output : 30
If P is greater than or equal to K then all problems will be solved on that day or (K-P) problems will be solved on each day so the answer will be 0 if K<=P else the answer will be (K-P)*N.
Below is the implementation of the above approach:
C++
// C++ program to find problems not // solved at the end of Nth day #include <bits/stdc++.h> using namespace std;
// Function to find problems not // solved at the end of Nth day int problemsLeft( int K, int P, int N)
{ if (K <= P)
return 0;
else
return (K - P) * N;
} // Driver Code int main()
{ int K, P, N;
K = 4;
P = 1;
N = 10;
cout << problemsLeft(K, P, N);
return 0;
} |
Java
// Java program to find problems not // solved at the end of Nth day class Gfg {
// Function to find problems not
// solved at the end of Nth day
public static int problemsLeft( int K, int P, int N)
{
if (K <= P)
return 0 ;
else
return ((K - P) * N);
}
// Driver Code
public static void main(String args[])
{
int K, P, N;
K = 4 ;
P = 1 ;
N = 10 ;
System.out.println(problemsLeft(K, P, N));
}
} |
Python3
# Python program to find problems not # solved at the end of Nth day def problemsLeft(K, P, N):
if (K< = P):
return 0
else :
return ((K - P) * N)
# Driver Code K, P, N = 4 , 1 , 10
print (problemsLeft(K, P, N))
|
C#
// C# program to find problems not // solved at the end of Nth day using System;
class GFG
{ // Function to find problems not // solved at the end of Nth day public static int problemsLeft( int K,
int P, int N)
{ if (K <= P)
return 0;
else
return ((K - P) * N);
} // Driver Code public static void Main()
{ int K, P, N;
K = 4;
P = 1;
N = 10;
Console.WriteLine(problemsLeft(K, P, N));
} } // This code is contributed by vt_m |
PHP
<?php // PHP program to find problems not // solved at the end of Nth day // Function to find problems not // solved at the end of Nth day function problemsLeft( $K , $P , $N )
{ if ( $K <= $P )
return 0;
else
return ( $K - $P ) * $N ;
} // Driver Code $K = 4;
$P = 1;
$N = 10;
echo problemsLeft( $K , $P , $N );
// This code is contributed by anuj_67 ?> |
Javascript
<script> // JavaScript program to find problems not // solved at the end of Nth day // Function to find problems not // solved at the end of Nth day function problemsLeft( K, P, N){
if (K <= P)
return 0;
else
return (K - P) * N;
} // Driver Code let K, P, N; K = 4;
P = 1;
N = 10;
document.write(problemsLeft(K, P, N)); // This code is contributed by rohitsingh07052. </script> |
Output:
30
Time Complexity: O(1)
Auxiliary Space: O(1)