Find the number after successive division
Last Updated :
08 Sep, 2022
Given two arrays div and rem which hold the values for divisors and remainders, the task is to find the number which after being successively divided by the elements of the div array leaves remainder which are in the rem array.
Note: The quotient from the first division will be divided by the second element and then the resultant quotient will be divided by the third (matching the remainders given) and so on.
Examples:
Input: div[] = {3, 5, 7}, rem[] = {1, 3, 5}
Output: 85
85 on division with 3 leaves remainder 1 with quotient 28
28 on division with 5 leaves remainder 3 with quotient 5
5 on division with 7 leaves remainder 5
Input: div[] = {7, 9}, rem[] = {2, 2}
Output: 16
Approach:
- Store the value of the last remainder in a variable say num.
- Traverse the array backwards from n-2 to 0 and update the num as num = num * div[i] + rem[i]
- Print num in the end.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findNum( int div [], int rem[], int N)
{
int num = rem[N - 1];
for ( int i = N - 2; i >= 0; i--) {
num = num * div [i] + rem[i];
}
return num;
}
int main()
{
int div [] = { 8, 3 };
int rem[] = { 2, 2 };
int N = sizeof ( div ) / sizeof ( div [0]);
cout << findNum( div , rem, N);
return 0;
}
|
Java
public class GFG{
static int findNum( int div[], int rem[], int N)
{
int num = rem[N - 1 ];
for ( int i = N - 2 ; i >= 0 ; i--) {
num = num * div[i] + rem[i];
}
return num;
}
public static void main(String []args){
int div[] = { 8 , 3 };
int rem[] = { 2 , 2 };
int N = div.length;
System.out.println(findNum(div, rem, N));
}
}
|
Python3
def findNum(div, rem, N):
num = rem[N - 1 ]
i = N - 2
while (i > = 0 ):
num = num * div[i] + rem[i]
i - = 1
return num
if __name__ = = '__main__' :
div = [ 8 , 3 ]
rem = [ 2 , 2 ]
N = len (div)
print (findNum(div, rem, N))
|
C#
using System;
class GFG
{
static int findNum( int []div,
int []rem, int N)
{
int num = rem[N - 1];
for ( int i = N - 2; i >= 0; i--)
{
num = num * div[i] + rem[i];
}
return num;
}
static public void Main ()
{
int []div = { 8, 3 };
int []rem = { 2, 2 };
int N = div.Length;
Console.WriteLine(findNum(div, rem, N));
}
}
|
PHP
<?php
function findNum( $div , $rem , $N )
{
$num = $rem [ $N - 1];
for ( $i = $N - 2; $i >= 0; $i --)
{
$num = $num * $div [ $i ] + $rem [ $i ];
}
return $num ;
}
$div = array ( 8, 3 );
$rem = array (2, 2 );
$N = sizeof( $div );
echo findNum( $div , $rem , $N );
?>
|
Javascript
<script>
function findNum(div, rem, N)
{
var num = rem[N - 1];
for ( var i = N - 2; i >= 0; i--) {
num = num * div[i] + rem[i];
}
return num;
}
var div = [ 8, 3 ];
var rem = [ 2, 2 ];
var N = div.length;
document.write( findNum(div, rem, N));
</script>
|
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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