URI Online Judge | 1018 Banknotes (Solution)

Problem Link:
https://www.urionlinejudge.com.br/judge/en/problems/view/1018

URI Online Judge | 1018

Banknotes

Adapted by Neilor Tonin, URI  Brazil
Timelimit: 1
In this problem you have to read an integer number and calculate the smallest possible number of notes in which the value may be decomposed. The notes are of 100, 50, 20, 10, 5, 2 e 1. Print the value read and the list of notes.

Input

The input file contain an integer number N (0 < N < 1000000).

Output

Print the read number and the minimum quantity of each necessary banknotes, like as the given example. Do not forget to print the end line after each line, otherwise you will get “Presentation Error”.
Sample InputSample Output
576576
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
1 nota(s) de R$ 1,00
1125711257
112 nota(s) de R$ 100,00
1 nota(s) de R$ 50,00
0 nota(s) de R$ 20,00
0 nota(s) de R$ 10,00
1 nota(s) de R$ 5,00
1 nota(s) de R$ 2,00
0 nota(s) de R$ 1,00
503503
5 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
1 nota(s) de R$ 2,00
1 nota(s) de R$ 1,00
Solution:

#include <stdio.h>
  
int main() {
  
int d[] = {100, 50, 20, 10, 5, 2, 1};
int n, t = 0;
scanf("%d", &n);
printf("%d\n", n);
do {
printf("%d nota(s) de R$ %d,00\n", n / d[t], d[t]);
if (n >= d[t])
n -= d[t] * (n / d[t]);
} while(d[t++] != 1);
  
    return 0;
}

Comments

  1. craigkillerUSAMarch 5, 2016 at 10:07 AM

    Beautiful

    ReplyDelete
  2. UnknownJune 2, 2016 at 10:25 PM

    This comment has been removed by the author.

    ReplyDelete
  3. UnknownJune 2, 2016 at 10:26 PM

    This comment has been removed by the author.

    ReplyDelete
  4. UnknownJune 2, 2016 at 10:27 PM

    #include

    int main() {

    int a,b,c,d,e,f,g;
    scanf("%d",&a);
    printf("%d",a);
    b = a%100;
    c = b%50;
    d = c%20;
    e = d%10;
    f = e%5;
    g = f%2;
    printf("%d nota(s) de R$ 100,00\n%d nota(s) de R$ 50,00\n%d nota(s) de R$ 20,00\n%d nota(s) de R$ 10,00\n%d nota(s) de R$ 5,00\n%d nota(s) de R$ 2,00\n%d nota(s) de R$ 1,00\n",a/100,b/50,c/20,d/10,e/5,f/2,g/1);

    return 0;
    }

    why this solve is showing 20% wrong answer?

    ReplyDelete
    Replies
    1. UnknownAugust 7, 2016 at 9:12 AM

      You should insert a \n in very first printf function. :)

      Delete
    2. AnonymousNovember 20, 2017 at 10:07 PM

      #include
      int main(void)
      {
      int i,j,n;
      int a[7]={100,50,20,10,5,2,1};
      int b[7];
      scanf("%d",&n);
      printf("%d\n",n);
      for(i=0;i<7;i++){
      b[i]=n/a[i];
      n=n%a[i];
      }
      for(j=0;j<7;j++){
      printf("%d nota(s) de R$ %d,00\n",b[j],a[j]);
      }
      return 0;
      }

      Delete
    3. UnknownNovember 5, 2018 at 12:35 PM

      this is compilation error but why...?
      i guess that is rigt.

      Delete
  5. AdminAugust 10, 2016 at 3:11 PM

    import java.util.Scanner;


    public class Main {

    public static void main(String[] args) {

    int taka;
    int a;
    int b;
    int c;
    int d;
    int e;
    int f;
    int g;
    int h;
    int i;
    int j;
    int k;
    int l;
    int m;



    Scanner input = new Scanner(System.in);

    taka = input.nextInt();

    if(taka > 0 || taka <1000000)
    {
    a = taka / 100;
    b = taka % 100;
    c = b / 50;
    d = b % 50;
    e = d / 20;
    f = d % 20;
    g = f / 10;
    h = f % 10;
    i = h /5;
    j = h % 5;
    k = j / 2;
    l = j % 2;
    m = l / 1;



    System.out.printf("%d nota(s) de R$ 100.00\n", a);
    System.out.printf("%d nota(s) de R$ 50.00\n", c);
    System.out.printf("%d nota(s) de R$ 20.00\n", e);
    System.out.printf("%d nota(s) de R$ 10.00\n", g);
    System.out.printf("%d nota(s) de R$ 5.00\n", i);
    System.out.printf("%d nota(s) de R$ 2.00\n", k);
    System.out.printf("%d nota(s) de R$ 1.00\n", m);

    }
    }

    }

    why i have seen 90% wrong answer??

    ReplyDelete
  6. UnknownNovember 7, 2016 at 9:27 AM

    int m=1 , n=2
    for(j=1 ;j<=2 ;j+1)
    {m=m+1;
    n=n*j;
    printf("%d\t %d\t" , m,n,);

    ReplyDelete
  7. Gulam Kibria(GK)May 23, 2017 at 8:03 AM

    #include
    int main()
    {
    int N,m,n,o,p,q,r,s;
    scanf("%d",&N);
    m=N/100;
    n=(N%100)/50;
    o=((N%100)%50)/20;
    p=(((N%100)%50)%20)/10;
    q=((((N%100)%50)%20)%10)/5;
    r=(((((N%100)%50)%20)%10)%5)/2;
    s=((((((N%100)%50)%20)%10)%5)%2)/1;
    printf("%d nota(s) de R$ 100,00\n%d nota(s) de R$ 50,00\n%d nota(s) de R$ 20,00\n%d nota(s) de R$ 10,00\n%d nota(s) de R$ 5,00\n%d nota(s) de R$ 2,00\n%d nota(s) de R$ 1,00\n",m,n,o,p,q,r,s);
    return 0;
    }




    wrong 100%.....but why.

    ReplyDelete
  8. UnknownJune 5, 2017 at 3:16 AM

    #include

    int main()
    {
    int N,A,B,C,D,E,F,G;

    scanf("%d",&N);

    if(0<N && N<1000000)
    {
    A=N/100;

    B=N%100/50;

    C=N%100%50/20;

    D=N%100%50%20/10;

    E=N%100%50%20%10/5;

    F=N%100%50%20%10%5/2;

    G=N%100%50%20%10%5%2/1;

    printf("%d nota(s) de R$ 100,00\n%d nota(s) de R$ 50,00\n%d nota(s) de R$ 20,00\n%d nota(s) de R$ 10,00\n%d nota(s) de R$ 5,00\n%d nota(s) de R$ 2,00\n%d nota(s) de R$ 1,00\n",A,B,C,D,E,F,G);
    }
    return 0;
    }

    ReplyDelete
  9. UnknownJuly 9, 2017 at 9:49 AM

    import java.util.Scanner;

    public class Main {

    public static void main(String[] args) {
    int givenbValue;
    Scanner scan = new Scanner(System.in);
    givenbValue = scan.nextInt();
    int fiftyTaka = givenbValue % 100;
    int hundredValue = givenbValue / 100;
    System.out.println(hundredValue + " nota(s) de R$ 100,00");
    int twentyTaka = fiftyTaka % 50;
    fiftyTaka = fiftyTaka / 50;
    System.out.println(fiftyTaka + " nota(s) de R$ 50,00");
    int tenTaka = twentyTaka % 20;
    twentyTaka = twentyTaka / 20;
    System.out.println(twentyTaka + " nota(s) de R$ 20,00");
    int fiveTaka = tenTaka % 10;
    tenTaka = tenTaka / 10;
    System.out.println(tenTaka + " nota(s) de R$ 10,00");
    int twoTaka = fiveTaka % 5;
    fiveTaka = fiveTaka / 5;
    System.out.println(fiveTaka + " nota(s) de R$ 5,00");
    int oneTaka = twoTaka % 2;
    twoTaka = twoTaka / 2;
    System.out.println(twoTaka + " nota(s) de R$ 2,00");
    oneTaka = oneTaka / 1;
    System.out.println(oneTaka + " nota(s) de R$ 1,00");


    }
    }

    ReplyDelete
  10. UnknownJuly 18, 2017 at 1:49 AM

    #include
    using namespace std;

    int main()
    {
    int n , a , b , c , d , e , f , g ;
    cin>>n;
    if (n>0 && n<1000000)
    {
    a=n/100;

    b=n%100/50;

    c=n%100%50/20;

    d=n%100%50%20/10;

    e=n%100%50%20%10/5;

    f=n%100%50%20%10%5/2;

    g=n%100%50%20%10%5%2/1;

    cout<<n<<endl;;
    cout<<a<<" nota(s) de R$ 100,00"<<endl;
    cout<<b<<" nota(s) de R$ 50,00"<<endl;
    cout<<c<<" nota(s) de R$ 20,00"<<endl;
    cout<<d<<" nota(s) de R$ 10,00"<<endl;
    cout<<e<<" nota(s) de R$ 5,00"<<endl;
    cout<<f<<" nota(s) de R$ 2,00"<<endl;
    cout<<g<<" nota(s) de R$ 1,00"<<endl;
    }

    return 0;
    }

    ReplyDelete

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