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Count the number of currency notes needed in C++
Given an amount in rupees that one has to pay, say pay_rupees and an unlimited amount of banknotes that have value as Rupees_amount_1 and Rupees_amount_2. The goal is to pay pay_rupees using exactly the total number of notes=distribution_total and count the number of type Rupees_amount_1 notes required. If there is no solution to pay then return −1 as answer.
For Example
Input
Rupees_amount_1 = 1, Rupees_amount_2 = 5, pay_Rupees = 11 distribution_total = 7
Output
Count of number of currency notes needed are − 6
Explanation
6*1 + 5*1 = 11 and notes=6+1=7
Input
Rupees_amount_1 = 2, Rupees_amount_2 = 3, pay_Rupees = 10 distribution_total = 4
Output
Count of number of currency notes needed are: 2
Explanation
2*2 + 3*2 = 10 and notes=2+2=4
Approach used in the below program is as follows −
Let a1 be the number of notes of amount rupees 1and N be the total number of notes. To make the payment of P, we will have equations − a1*(Rupees_amount_1) + (N−a1)*(Rupees_amount_2) = P P=a1*(Rupees_amount_1) + (N)*(Rupees_amount_2) − (a1)*(Rupees_amount_2) P − (N)*(Rupees_amount_2) =a1*(Rupees_amount_1) − (a1)*(Rupees_amount_2) a1= P − (N)*(Rupees_amount_2) / ( Rupees_amount_1 − Rupees_amount_2) If a1 is integer then only we will have a solution else return −1.
Take all numbers as input.
Function notes_needed(int Rupees_amount_1, int Rupees_amount_2, int pay_Rupees, int distribution_total) takes all inputs and returns the count of the number of currency notes needed.
Take the initial count as 0.
Take total = pay_Rupees − (Rupees_amount_2 * distribution_total)
Set total_given = Rupees_amount_1 − Rupees_amount_2
Set total_given = Rupees_amount_1 − Rupees_amount_2
If total % total_given == 0 then return count as total / total_given.
Else return −1.
Example
#include<bits/stdc++.h> using namespace std; int notes_needed(int Rupees_amount_1, int Rupees_amount_2, int pay_Rupees, int distribution_total){ int count = 0; int total = pay_Rupees − (Rupees_amount_2 * distribution_total); int total_given = Rupees_amount_1 − Rupees_amount_2; if (total % total_given == 0){ count = total / total_given; return count; } else { return −1; } } int main(){ int Rupees_amount_1 = 1; int Rupees_amount_2 = 5; int pay_Rupees = 11; int distribution_total = 7; cout<<"Count of number of currency notes needed are: "<<notes_needed(Rupees_amount_1, Rupees_amount_2, pay_Rupees, distribution_total); }
Output
If we run the above code it will generate the following output −
Count the number of currency notes needed are− 6