Saturday, 16 January 2016

Solution of URI 1021 :: Banknotes and Coins

do you feel your code Basic Code, C, URI 15 comments

Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
int main()
{
    double A,D,E;
    scanf ("%lf",&A);
    int N,a,b,c,d,e,f,g,h,i,j,k,l,B,m,n,o,p,q,r,s,t,u;
    N=A;
    a=N/100;
    b=N%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;

    E=A*100;
    B=(int) E;
    m=B%100;
    n=m/50;
    o=m%50;
    p=o/25;
    q=o%25;
    r=q/10;
    s=q%10;
    t=s/5;
    u=s%5;

    printf ("NOTAS:\n");
    printf ("%d nota(s) de R$ 100.00\n",a);
    printf ("%d nota(s) de R$ 50.00\n",c);
    printf ("%d nota(s) de R$ 20.00\n",e);
    printf ("%d nota(s) de R$ 10.00\n",g);
    printf ("%d nota(s) de R$ 5.00\n",i);
    printf ("%d nota(s) de R$ 2.00\n",k);
    printf ("MOEDAS:\n");
    printf ("%d moeda(s) de R$ 1.00\n",l);
    printf ("%d moeda(s) de R$ 0.50\n",n);
    printf ("%d moeda(s) de R$ 0.25\n",p);
    printf ("%d moeda(s) de R$ 0.10\n",r);
    printf ("%d moeda(s) de R$ 0.05\n",t);
    printf ("%d moeda(s) de R$ 0.01\n",u);

    return 0;
}

15 comments:

mb-billah_18 March 2016 at 21:25
This comment has been removed by a blog administrator.
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Crítico cabra da peste15 August 2016 at 22:17
Puts cara, eu fiz em python mas não esta aceitando :(. Esta batendo certinho, mas fala que esta errado...
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Crítico cabra da peste15 August 2016 at 22:17
Puts cara, eu fiz em python mas não esta aceitando :(. Esta batendo certinho, mas fala que esta errado...
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Anonymous26 January 2017 at 20:41
por acaso da wrong answer 10% ?
o meu ta a mesma coisa. bate tudo mas nao aceita
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Anonymous23 May 2017 at 21:08
GALERA ALGUEM SABE QUAL SERIA O PROBLEMA DO MEU CODIGO ESTOU COM 10% DE ERRO!!

#include

int main() {

int n, m;
scanf("%d.%d", &n, &m);

printf("NOTAS:\n%d nota(s) de R$ 100.00\n", n/100);
printf("%d nota(s) de R$ 50.00\n", n%100/50);
printf("%d nota(s) de R$ 20.00\n", n%100%50/20);
printf("%d nota(s) de R$ 10.00\n", n%100%50%20/10);
printf("%d nota(s) de R$ 5.00\n", n%100%50%20%10/5);
printf("%d nota(s) de R$ 2.00\n", n%100%50%20%10%5/2);

printf("MOEDAS:\n%d moeda(s) de R$ 1,00\n", n%100%50%20%10%5%2);
printf("%d moeda(s) de R$ 0.50\n", m%100/50);
printf("%d moeda(s) de R$ 0.25\n", m%100%50/25);
printf("%d moeda(s) de R$ 0.10\n", m%100%50%25/10);
printf("%d moeda(s) de R$ 0.05\n", m%100%50%25%10/5);
printf("%d moeda(s) de RS 0.01\n", m%100%50%25%10%5);

return 0;
}
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Unknown6 October 2017 at 06:11
Quando uso 576.43 como entrada, o programa só conta 2 moedas de 1 centavo, não entendo o pq.

#include

int main(int argc, char **argv)
{
float valor, cedulas[6]={100.00, 50.00, 20.00, 10.00, 5.00, 2.00}, moedas[6]={1.00, 0.50, 0.25, 0.10, 0.05, 0.01};
int notas,i, quantiMoedas;

scanf("%f", &valor);

printf("NOTAS:\n");
i=0;

while(i < 6){
notas = valor / cedulas[i];
printf("%d nota(s) de R$ %1.2f\n", notas, cedulas[i]);
valor -= (cedulas[i] * notas);
i++;
}

i=0;

printf("MOEDAS:\n");
while(i < 6){
quantiMoedas = valor / moedas[i];
printf("%d moeda(s) de R$ %1.2f\n", quantiMoedas, moedas[i]);
valor -= (moedas[i] * quantiMoedas);
i++;
}

return 0;
}

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Anonymous17 January 2018 at 10:56
código funciona em todos os casos de teste mas me da WA 10%,não da para entender...
float n;
int valor,inicio,resto;
scanf("%f", &n);
valor=n;
inicio=valor;
printf("NOTAS:\n%d nota(s) de R$ 100.00\n",valor/100);
valor=valor%100;
printf("%d nota(s) de R$ 50.00\n",valor/50);
valor=valor%50;
printf("%d nota(s) de R$ 20.00\n",valor/20);
valor=valor%20;
printf("%d nota(s) de R$ 10.00\n",valor/10);
valor=valor%10;
printf("%d nota(s) de R$ 5.00\n",valor/5);
valor=valor%5;
printf("%d nota(s) de R$ 2.00\n",valor/2);
valor=valor%2;
printf("MOEDAS:\n%d moeda(s) de R$ 1.00\n",valor);
resto=(n*100)-(inicio*100)+0.01;
printf("%d moeda(s) de R$0.50\n",resto/50);
resto=resto%50;
printf("%d moeda(s) de R$0.25\n",resto/25);
resto=resto%25;
printf("%d moeda(s) de R$0.10\n",resto/10);
resto=resto%10;
printf("%d moeda(s) de R$0.05\n",resto/5);
resto=resto%5;
printf("%d moeda(s) de R$0.01\n",resto);
return 0;
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Eterno Aprendiz20 April 2018 at 20:17
Where's the error? (Onde está o erro?)

Language: C++

#include

using namespace std;

int main()
{
double valor, valores_moeda[12] = {100,50,20,10,5,2,1,0.50,0.25,0.10,0.05,0.01};
int i, valor_decomposto, quant_moedas[12] = {0};
string texto_moedas[12] = {"100.00","50.00","20.00","10.00","5.00","2.00",
"1.00","0.50","0.25","0.10","0.05","0.01"};

cin >> valor;

for(i = 0; i < 12; i++)
{
valor_decomposto = valor / valores_moeda[i];
quant_moedas[i] = valor_decomposto;
valor = valor - (valores_moeda[i] * (double) valor_decomposto);

if(i==0)
cout << "NOTAS:" << endl;

if(i==6)
cout << "MOEDAS:" << endl;

if(i<6)
cout << quant_moedas[i] << " nota(s) de R$ " << texto_moedas[i] << endl;
else
cout << quant_moedas[i] << " moeda(s) de R$ " << texto_moedas[i] << endl;
}

return 0;
}
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Unknown5 July 2018 at 15:36
for URI 1021

when you approach to find the number of moedas , then
just you have to do the little this to avoid 10% / 20% error ALTHOUGH YOUR ANSWER WAS CORRECT BUT NOT ACCEPTED BY URI COMPILER
N =576.73.
int M = (int)N; // M= 576
float O = N-M; // O = .72922 *** problem is here***
int I = (int)O*100; //I = 72 but we took .73
M= (int) O * 1000; // M = 729
if ( M % 10 > = 5 )
{ I++; }
// now you print example
int ft= i/50;

printf( "%d moeda(s) de R$ 0.50\n",FT);
ft= i%50/25;
printf( "%d moeda(s) de R$ 0.25\n",FT);
ft= i%50%25/10;
printf( "%d moeda(s) de R$ 0.10\n",ft);
ft= i%50%25%10/5;

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