URI Online Judge Solution 1021 Banknotes and Coins – URI 1021 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1021 Banknotes and Coins – URI 1021 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1021 Banknotes and Coins | Beginner
URI Problem Link – URI 1021 Banknotes and Coins Problem – https://www.urionlinejudge.com.br/judge/en/problems/view/1021

Problem Name: 1021 Banknotes and Coins code
Problem Number : URI – 1021 Banknotes and Coins solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1021 Banknotes and Coins Code in C / URI 1021 solution in C:

``#include <stdio.h>int main() {    int n100, n50, n20, n10, n5, n2;    int m1, m50, m25, m10, m05, m01;    double n;    scanf("%lf", &n);    int notas = n;    int moedas = (n - notas) * 100;    if((moedas * 1000) % 10 == 9){        moedas++;    }    n100 = notas/100;    notas = notas%100;    n50 = notas/50;    notas = notas%50;    n20 = notas/20;    notas = notas%20;    n10 = notas/10;    notas = notas%10;    n5 = notas/5;    notas = notas%5;    n2 = notas/2;    notas = notas%2;    m1 = notas/1;    notas = notas%1;    m50 = moedas/50;    moedas = moedas%50;    m25 = moedas/25;    moedas = moedas%25;    m10 = moedas/10;    moedas = moedas%10;    m05 = moedas/5;    moedas = moedas%5;    m01 = moedas/1;    printf("NOTAS:n");    printf("%d nota(s) de R\$ 100.00n", n100);    printf("%d nota(s) de R\$ 50.00n", n50);    printf("%d nota(s) de R\$ 20.00n", n20);    printf("%d nota(s) de R\$ 10.00n", n10);    printf("%d nota(s) de R\$ 5.00n", n5);    printf("%d nota(s) de R\$ 2.00n", n2);    printf("MOEDAS:n");    printf("%d moeda(s) de R\$ 1.00n", m1);    printf("%d moeda(s) de R\$ 0.50n", m50);    printf("%d moeda(s) de R\$ 0.25n", m25);    printf("%d moeda(s) de R\$ 0.10n", m10);    printf("%d moeda(s) de R\$ 0.05n", m05);    printf("%d moeda(s) de R\$ 0.01n", m01);    return 0;}``

URI Solution 1021 Banknotes and Coins Code / URI 1002 solution in CPP:

``#include <iostream>using namespace std;int main() {    double n100, n50, n20, n10, n5, n2;    double m1, m50, m25, m10, m05, m01;    double n;    cin >> n;    int notas = n;    int moedas = (n - notas) * 100;    if((moedas * 1000) % 10 == 9){        moedas++;    }    n100 = notas/100;    notas = notas%100;    n50 = notas/50;    notas = notas%50;    n20 = notas/20;    notas = notas%20;    n10 = notas/10;    notas = notas%10;    n5 = notas/5;    notas = notas%5;    n2 = notas/2;    notas = notas%2;    m1 = notas/1;    notas = notas%1;    m50 = moedas/50;    moedas = moedas%50;    m25 = moedas/25;    moedas = moedas%25;    m10 = moedas/10;    moedas = moedas%10;    m05 = moedas/5;    moedas = moedas%5;    m01 = moedas/1;    cout << "NOTAS:" << endl;     cout << n100 << " nota(s) de R\$ 100.00" << endl;    cout << n50 << " nota(s) de R\$ 50.00" << endl;    cout << n20 << " nota(s) de R\$ 20.00" << endl;    cout << n10 << " nota(s) de R\$ 10.00" << endl;    cout << n5 << " nota(s) de R\$ 5.00" << endl;    cout << n2 << " nota(s) de R\$ 2.00" << endl;    cout << "MOEDAS:" << endl;    cout << m1 << " moeda(s) de R\$ 1.00" << endl;    cout << m50 << " moeda(s) de R\$ 0.50" << endl;    cout << m25 << " moeda(s) de R\$ 0.25" << endl;    cout << m10 << " moeda(s) de R\$ 0.10" << endl;    cout << m05 << " moeda(s) de R\$ 0.05" << endl;    cout << m01 << " moeda(s) de R\$ 0.01" << endl;    return 0;}``

URI Solution 1021 Banknotes and Coins Code / URI 1021 Banknotes and Coins solution in Java:

``import java.util.Scanner;public class Main{    public static void main(String[] args)    {        float x;        int note100,note50,note20,note10,note5,note2;        int moeda1,moeda5,moeda25,moeda10,moeda05,moeda01;        int reminder100;        Scanner input=new Scanner(System.in);        x = input.nextFloat();        note100 =(int) x / 100;        reminder100 = (int) (x % 100);        note50 = (reminder100) / 50;        note20 = (reminder100 % 50 )/ 20;        note10 = ((reminder100 % 50 )% 20) / 10;        note5 = (((reminder100 % 50 )% 20) % 10) / 5;        note2 = (((reminder100 % 50 )% 20) % 5) / 2;        //------     MOEDAS:    ------------//        moeda1 =  (((((reminder100 % 50 )% 20) % 5) % 2) / 1);        float reminderMoeda = (float) (((((reminder100 % 50 )% 20) % 5) % 2));        float meda5Float = (float) ((reminderMoeda % 1) / .5);        moeda5 = (int) (meda5Float);        moeda25 = (int) (((((((reminder100 % 50 )% 20) % 5) % 2) % 1) % .5) / .25);        moeda10 = (int) ((((((((reminder100 % 50 )% 20) % 5) % 2) % 1) % .5) % .25) / .1);        moeda05 = (int) (((((((((reminder100 % 50 )% 20) % 5) % 2) % 1) % .5) % .25) % .1) / .05);        moeda01 = (int) ((((((((((reminder100 % 50 )% 20) % 5) % 2) % 1) % .5) % .25) % .1) % .05) / .01);        System.out.print("NOTAS:n");        System.out.print(note100 +" nota(s) de R\$ 100.00n");        System.out.print(note50 +" nota(s) de R\$ 50.00n");        System.out.print(note20 +" nota(s) de R\$ 20.00n");        System.out.print(note10 +" nota(s) de R\$ 10.00n");        System.out.print(note5 +" nota(s) de R\$ 5.00n");        System.out.print(note2 +" nota(s) de R\$ 2.00nn");        System.out.print("MOEDAS:n");        System.out.print(moeda1 +" moeda(s) de R\$ 1.00n");        System.out.print(meda5Float +" moeda(s) de R\$ 0.50n");        System.out.print(moeda25 +" moeda(s) de R\$ 0.25n");        System.out.print(moeda10 +" moeda(s) de R\$ 0.10n");        System.out.print(moeda05 +" moeda(s) de R\$ 0.05n");        System.out.print(moeda01 +" moeda(s) de R\$ 0.01n");    }}``

URI Solution 1021Code / URI 1021 Banknotes and Coins solution in  Python:

``value = eval(input());cem = cinquenta = vinte = dez = cinco = dois = um = 0;cincents = vintecincents = dezcents = cincocents = cents = 0;value = float("%.2f" % value)if int(value/100) >= 1: cem = int(value/100); value -= cem*100;value = float("%.2f" % value)if int(value/50) >= 1: cinquenta = int(value/50); value -= cinquenta*50;value = float("%.2f" % value)if int(value/20) >= 1: vinte = int(value/20.00); value -= vinte*20;value = float("%.2f" % value)if int(value/10) >= 1: dez = int(value/10); value -= dez*10.00;value = float("%.2f" % value)if int(value/5) >= 1: cinco = int(value/5); value -= cinco*5;value = float("%.2f" % value)if int(value/2) >= 1: dois = int(value/2); value -= dois*2;value = float("%.2f" % value)if int(value/1) >= 1: um = int(value/1); value -= um*1;value = float("%.2f" % value)if int(value/0.50) >= 1: cincents = int(value/0.50); value -= cincents*0.50;value = float("%.2f" % value)if int(value/0.25) >= 1: vintecincents = int(value/0.25); value -= vintecincents*0.25;value = float("%.2f" % value)if int(value/0.10) >= 1: dezcents = int(value/0.10); value -= dezcents*0.10;value = float("%.2f" % value)if int(value/0.05) >= 1: cincocents = int(value/0.05); value -= cincocents*0.05;value = float("%.2f" % value)if int(value/0.01) >= 0.998: cents = int(value/0.01); value -= cents*0.01;print("NOTAS:");print("%d nota(s) de R\$ 100.00" % cem);print("%d nota(s) de R\$ 50.00" % cinquenta);print("%d nota(s) de R\$ 20.00" % vinte);print("%d nota(s) de R\$ 10.00" % dez);print("%d nota(s) de R\$ 5.00" % cinco);print("%d nota(s) de R\$ 2.00" % dois);print("MOEDAS:");print("%d moeda(s) de R\$ 1.00" % um);print("%d moeda(s) de R\$ 0.50" % cincents);print("%d moeda(s) de R\$ 0.25" % vintecincents);print("%d moeda(s) de R\$ 0.10" % dezcents);print("%d moeda(s) de R\$ 0.05" % cincocents);print("%d moeda(s) de R\$ 0.01" % cents);``

URI Solution 1021Code / URI 1021 Banknotes and Coins solution in  C# (C Sharp):

Demonstration:

It’s a problem of URI online Judge to convert notes. Use reminder to get the notes part and try..
Just implement this in coding. Since having any problem just put a comment below. Thanks

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By Maniruzzaman Akash

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