URI Online Judge | 1021 Banknotes and Coins (Solution)

Problem Link:
https://www.urionlinejudge.com.br/judge/en/problems/view/1021

URI Online Judge | 1021

Banknotes and Coins

By Neilor Tonin, URI  Brazil
Timelimit: 1
Read a floating point value with two decimal places. This represents a monetary value. Next, calculate the fewest possible notes and coins in which the value can be decomposed. The considered notes are of 100, 50, 20, 10, 5, 2. The possible coins are of 1, 0.50, 0.25, 0.10, 0.05 and 0.01. Print the message “NOTAS:” followed by the list of notes and the message “MOEDAS:” followed by the list of coins.

Input

The input file contain float point number N (0 ≤ N ≤ 1000000.00).

Output

Print the minimum quantity of banknotes and coins necessary to change the initial value, like as the given example.
Sample InputSample Output
576.73NOTAS:
5 nota(s) de R$ 100.00
1 nota(s) de R$ 50.00
1 nota(s) de R$ 20.00
0 nota(s) de R$ 10.00
1 nota(s) de R$ 5.00
0 nota(s) de R$ 2.00
MOEDAS:
1 moeda(s) de R$ 1.00
1 moeda(s) de R$ 0.50
0 moeda(s) de R$ 0.25
2 moeda(s) de R$ 0.10
0 moeda(s) de R$ 0.05
3 moeda(s) de R$ 0.01
4.00NOTAS:
0 nota(s) de R$ 100.00
0 nota(s) de R$ 50.00
0 nota(s) de R$ 20.00
0 nota(s) de R$ 10.00
0 nota(s) de R$ 5.00
2 nota(s) de R$ 2.00
MOEDAS:
0 moeda(s) de R$ 1.00
0 moeda(s) de R$ 0.50
0 moeda(s) de R$ 0.25
0 moeda(s) de R$ 0.10
0 moeda(s) de R$ 0.05
0 moeda(s) de R$ 0.01
91.01NOTAS:
0 nota(s) de R$ 100.00
1 nota(s) de R$ 50.00
2 nota(s) de R$ 20.00
0 nota(s) de R$ 10.00
0 nota(s) de R$ 5.00
0 nota(s) de R$ 2.00
MOEDAS:
1 moeda(s) de R$ 1.00
0 moeda(s) de R$ 0.50
0 moeda(s) de R$ 0.25
0 moeda(s) de R$ 0.10
0 moeda(s) de R$ 0.05
1 moeda(s) de R$ 0.01
Solution:
#include <stdio.h>
int main()
{
    double n, d[] = {100.0, 50.0, 20.0, 10.0, 5.0, 2.0, 1.0, 0.5, 0.25, 0.10, 0.05, 0.01};
    int t = 0, c;
    scanf("%lf", &n);
    printf("NOTAS:\n");
    t = 0;
    n+=1e-9;
    while (d[t] >= 0.01)
    {
        c = 0;
        while (n >= d[t])
        {
            n -= d[t];
            c++;
        }
        if (d[t] == 1.0)
            printf("MOEDAS:\n");
        if (d[t] >= 2.0 )
            printf("%d nota(s) de R$ %.2f\n", c, d[t]);
        else
            printf("%d moeda(s) de R$ %.2f\n", c, d[t]);
        t++;
    }
    return 0;
}

Comments

  1. UnknownFebruary 3, 2016 at 12:07 PM

    hey, i'm your fan ;)

    ReplyDelete
  2. alpaMarch 22, 2016 at 6:31 PM

    Boa noite parabens pelo trabalho..so uma sugestao seria possivel voce comentar os codigos de maneira explicativa por gentileza

    ReplyDelete
  3. UnknownJune 8, 2016 at 9:16 AM

    This comment has been removed by the author.

    ReplyDelete
  4. UnknownJune 8, 2016 at 9:16 AM

    why it doesn't work in the first one :/
    #include

    using namespace std;
    int hund,fif,twent,ten,five,two,one;
    int num1,hund1,fif1,twent1,ten1,five1;
    float num;

    int main()
    {
    scanf("%f",&num);


    while(num-100>=0){
    num-=100;
    hund++;
    }
    while(num-50>=0){
    num-=50;
    fif++;
    }
    while(num-20>=0){
    num-=20;
    twent++;
    }
    while(num-10>=0){
    num-=10;
    ten++;
    }
    while(num-5>=0){
    num-=5;
    five++;
    }
    while(num-2>=0){
    num-=2;
    two++;
    }
    while(num-1>=0){
    num-=1;
    one++;
    }

    while(num-0.50>=0){
    num-=0.50;
    hund1++;
    }
    while(num-0.25>=0){
    num-=0.25;
    fif1++;
    }
    while(num-0.10>=0){
    num-=0.10;
    twent1++;
    }
    while(num-0.05>=0){
    num-=0.05;
    ten1++;
    }
    while(num-0.01>=0){
    num-=0.01;
    five1++;
    }



    printf("NOTAS:\n");
    printf("%d nota(s) de R$ 100.00\n",hund);
    printf("%d nota(s) de R$ 50.00\n",fif);
    printf("%d nota(s) de R$ 20.00\n",twent);
    printf("%d nota(s) de R$ 10.00\n",ten);
    printf("%d nota(s) de R$ 5.00\n",five);
    printf("%d nota(s) de R$ 2.00\n",two);
    printf("MOEDAS:\n");
    printf("%d moeda(s) de R$ 1.00\n",one);
    printf("%d moeda(s) de R$ 0.50\n",hund1);
    printf("%d moeda(s) de R$ 0.25\n",fif1);
    printf("%d moeda(s) de R$ 0.10\n",twent1);
    printf("%d moeda(s) de R$ 0.05\n",ten1);
    printf("%d moeda(s) de R$ 0.01\n",five1);


    return 0;
    }

    ReplyDelete
  5. UnknownOctober 26, 2016 at 1:25 AM

    Why my code is wrong for 1021 problem.....my code is..


    #include

    int main ()
    {
    double a;
    int b;

    scanf(" %lf", &a);
    printf("NOTAS:\n");

    b = a / 100;
    printf("%d nota(s) de R$ 100.00\n", b);
    a = a - ( b * 100);

    b = a / 50;
    printf("%d nota(s) de R$ 50.00\n", b);
    a = a - ( b * 50);

    b = a / 20;
    printf("%d nota(s) de R$ 20.00\n", b);
    a = a - ( b * 20);

    b = a / 10;
    printf("%d nota(s) de R$ 10.00\n", b);
    a = a - ( b * 10);

    b = a / 5;
    printf("%d nota(s) de R$ 5.00\n", b);
    a = a - ( b * 5);

    b = a / 2;
    printf("%d nota(s) de R$ 2.00\n", b);
    a = a - ( b * 2);

    printf("MOEDAS:\n");

    b = a / 1;
    printf("%d moeda(s) de R$ 1.00\n", b);
    a = a - ( b * 1);

    b = a / .50;
    printf("%d moeda(s) de R$ 0.50\n", b);
    a = a - ( b * .50);

    b = a / .25;
    printf("%d moeda(s) de R$ 0.25\n", b);
    a = a - ( b * .25);

    b = a / .10;
    printf("%d moeda(s) de R$ 0.10\n", b);
    a = a - ( b * .10);

    b = a / .05;
    printf("%d moeda(s) de R$ 0.05\n", b);
    a = a - ( b * .05);

    b = a / .01;
    printf("%d moeda(s) de R$ 0.01\n", b);


    return 0;
    }

    ReplyDelete
  6. AnnonJanuary 4, 2017 at 6:16 PM

    n+=1e-9;
    "1e-9" what do u mean by that??

    ReplyDelete
    Replies
    1. UnknownApril 10, 2017 at 11:50 AM

      that is exponent float declaration. Here base 10 is replaced by e Or E. we can write 1.2*10^-3 as 1.2e-3...

      Delete
  7. Zakaria KhanMarch 22, 2018 at 12:32 PM

    Why this is not accepting in JAVA? Showing 10% error.

    import java.util.Scanner;
    public class Main{
    public static void main(String[]args){
    Scanner sc=new Scanner(System.in);
    double N,rn,j;
    int n100=0,n50=0,n20=0,n10=0,n5=0,n2=0,p1=0,p50=0,p25=0,p10=0,p5=0,p01=0;
    N=sc.nextDouble();
    if (0<=N && N<= 1000000.00){
    j=N/100;
    n100=(int) j;
    rn=N%100;
    j=rn/50;
    n50=(int) j;
    rn=rn%50;
    j=rn/20;
    n20=(int) j;
    rn=rn%20;
    j=rn/10;
    n10=(int) j;
    rn=rn%10;
    j=rn/5;
    n5=(int) j;
    rn=rn%5;
    j=rn/2;
    n2=(int) j;
    rn=rn%2;
    j=rn/1;
    p1=(int) j;
    rn=rn%1;
    j=rn/0.5;
    p50=(int) j;
    rn=rn%0.5;
    j=rn/0.25;
    p25=(int) j;
    rn=rn%0.25;
    j=rn/0.10;
    p10=(int) j;
    rn=rn%0.10;
    j=rn/0.05;
    p5=(int) j;
    rn=rn%0.05;
    j=rn/0.01;
    p01=(int) j;
    }
    System.out.println("NOTAS:");
    System.out.println(n100+" nota(s) de R$ 100.00");
    System.out.println(n50+" nota(s) de R$ 50.00");
    System.out.println(n20+" nota(s) de R$ 20.00");
    System.out.println(n10+" nota(s) de R$ 10.00");
    System.out.println(n5+" nota(s) de R$ 5.00");
    System.out.println(n2+" nota(s) de R$ 2.00");
    System.out.println("MOEDAS:");
    System.out.println(p1+" moeda(s) de R$ 1.00");
    System.out.println(p50+" moeda(s) de R$ 0.50");
    System.out.println(p25+" moeda(s) de R$ 0.25");
    System.out.println(p10+" moeda(s) de R$ 0.10");
    System.out.println(p5+" moeda(s) de R$ 0.05");
    System.out.println(p01+" moeda(s) de R$ 0.01");
    }
    }

    ReplyDelete
  8. UnknownApril 22, 2019 at 7:38 AM

    Why is this not accepting in JAVA? Showing 10% error. Please, help me with this code.

    import java.io.IOException;
    import java.util.Scanner;
    public class Main {
    public static void main(String[] args) throws IOException {
    Scanner ler = new Scanner(System.in);
    double x;
    double moeda1=0, moeda2=0, moeda3=0, moeda4=0, moeda5=0, moeda6=0, nota=0, nota1=0, nota2=0, nota3=0, nota4=0, nota5=0, nota6=0;;
    x = ler.nextDouble();

    if (x >= 100) {
    nota = x / 100;
    nota = (int) x;
    x = x - (nota * 100);

    }
    if (x >= 50) {
    nota1 = x / 50;
    nota1 = (int) x;
    x = x - (nota1 * 50);

    }
    if (x >= 20) {
    nota2 = x / 20;
    x = x - (nota2 * 20);
    }
    if (x >= 10) {
    nota3 = x / 10;
    x = x - (nota3 * 10);
    }
    if (x >= 5) {
    nota4 = x / 5;
    x = x - (nota4 * 5);
    }
    if (x >= 2) {
    nota5 = x / 2;
    x = x - (nota5 * 2);
    }
    if (x >= 1) {
    moeda1 = x / 1;
    x = x - (moeda1 * 1);
    }
    if (x >= 0.50) {
    moeda2 = x / 0.50;
    x = x - (moeda2 * 0.50);
    }
    if (x >= 0.25) {
    moeda3 = x / 0.25;
    x = x - (moeda3 * 0.25);
    }
    if (x >= 0.10) {
    moeda4 = x / 0.10;
    x = x - (moeda4 * 0.10);
    }
    if (x >= 0.05) {
    moeda5 = x / 0.05;
    x = x - (moeda5 * 0.05);
    }
    if (x >= 0.01) {
    moeda6 = x / 0.01;
    x = x - (moeda6 * 0.01);
    }
    System.out.println("NOTAS:");
    System.out.println(nota + " nota(s) de R$ 100.00");
    System.out.println(nota1 + " nota(s) de R$ 50.00");
    System.out.println(nota2 + " nota(s) de R$ 20.00");
    System.out.println(nota3 + " nota(s) de R$ 10.00");
    System.out.println(nota4 + " nota(s) de R$ 5.00");
    System.out.println(nota5 + " nota(s) de R$ 2.00");
    System.out.println("MOEDAS:");
    System.out.println(moeda1 + " moeda(s) de R$ 1.00");
    System.out.println(moeda2 + " moeda(s) de R$ 0.50");
    System.out.println(moeda3 + " moeda(s) de R$ 0.25");
    System.out.println(moeda4 + " moeda(s) de R$ 0.10");
    System.out.println(moeda5 + " moeda(s) de R$ 0.05");
    System.out.println(moeda6 + " moeda(s) de R$ 0.01");
    }
    }

    ReplyDelete
  9. UnknownOctober 12, 2019 at 5:45 AM

    import java.util.Scanner;

    public class Main {

    public static void main(String[] args) {

    float a;
    int tmp;
    Scanner ssm = new Scanner(System.in);
    a=ssm.nextFloat();
    System.out.println("NOTAS:");
    tmp=(int) (a/100);
    System.out.printf("%d nota(s) de R$ 100.00\n",tmp);
    tmp=(int) ((a%100)/50);
    System.out.printf("%d nota(s) de R$ 50.00\n",tmp);
    tmp=(int) (((a%100)%50)/20);
    System.out.printf("%d nota(s) de R$ 20.00\n",tmp);
    tmp=(int) ((((a%100)%50)%20)/10);
    System.out.printf("%d nota(s) de R$ 10.00\n",tmp);
    tmp=(int) (((((a%100)%50)%20)%10)/5);
    System.out.printf("%d nota(s) de R$ 5.00\n",tmp);
    tmp=(int) ((((((a%100)%50)%20)%10)%5)/2);
    System.out.printf("%d nota(s) de R$ 2.00\n",tmp);
    System.out.println("MOEDAS:");
    tmp=(int) (((((((a%100)%50)%20)%10)%5)%2)/1);
    System.out.printf("%d moeda(s) de R$ 1.00\n",tmp);
    tmp=(int) ((float) (((((((a%100)%50)%20)%10)%5)%2)%1)/0.50);
    System.out.printf("%d moeda(s) de R$ 0.50\n",tmp);
    tmp=(int)((float) ((((((((a%100)%50)%20)%10)%5)%2)%1)%0.50)/0.25);
    System.out.printf("%d moeda(s) de R$ 0.25\n",tmp);
    tmp=(int) ((float) (((((((((a%100)%50)%20)%10)%5)%2)%1)%0.50)%0.25)/0.10);
    System.out.printf("%d moeda(s) de R$ 0.10\n",tmp);
    tmp=(int) ((float) ((((((((((a%100)%50)%20)%10)%5)%2)%1)%0.50)%0.25)%0.10)/0.05);
    System.out.printf("%d moeda(s) de R$ 0.05\n",tmp);
    tmp=(int) ((float) (((((((((((a%100)%50)%20)%10)%5)%2)%1)%0.50)%0.25)%0.10)%0.05)/0.01);
    System.out.printf("%d moeda(s) de R$ 0.01\n",tmp);
    }

    }


    why this program show wrong answer?

    ReplyDelete

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