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 dayint problemsLeft(int K, int P, int N){ if (K <= P) return 0; else return (K - P) * N;}// Driver Codeint 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 dayclass 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 daydef problemsLeft(K, P, N): if(K<= P): return 0 else: return ((K-P)*N)# Driver CodeK, P, N = 4, 1, 10print(problemsLeft(K, P, N)) |
C#
// C# program to find problems not// solved at the end of Nth dayusing System;class GFG {// Function to find problems not// solved at the end of Nth daypublic static int problemsLeft(int K, int P, int N){ if (K <= P) return 0; else return ((K - P) * N);}// Driver Codepublic 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 dayfunction 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 dayfunction problemsLeft( K, P, N){ if (K <= P) return 0; else return (K - P) * N;}// Driver Codelet 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)
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