Uri problem 1021 solve

Problem :

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 Input	Sample Output
576.73	NOTAS: 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

4.00	NOTAS: 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

91.01

NOTAS: 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() {

  

   int hundred ,fifty , twenty , ten ,five , two, coin1, coin2 ,coin3 ,coin4 ,coin5    ;

   float  input , coin6;

   scanf("%f",&input);

 

   hundred=(input/100);

   fifty=(input-(hundred*100))/50;

   twenty=(input-((fifty*50)+(hundred*100)))/20;

   ten=(input-((twenty*20)+(fifty*50)+(hundred*100)))/10;

   five=(input-((ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/5;

   two=(input-((five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/2;

 

   coin1=(input-((two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)));

   coin2=(input-((coin1*1.00)+(two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/0.5;

   coin3=(input-((coin2*0.5)+(coin1)+(two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/0.25;

   coin4=(input-((coin3*0.25)+(coin2*0.5)+(coin1)+(two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/0.10;

   coin5=(input-((coin4*0.10)+(coin3*.25)+(coin2*0.5)+(coin1)+(two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/0.05;

   coin6=(input-((coin5*0.05)+(coin4*.10)+(coin3*.25)+(coin2*0.5)+(coin1)+(two*2)+(five*5)+(ten*10)+(twenty*20)+(fifty*50)+(hundred*100)))/0.01;

   printf("NOTAS:\n");

   printf("%d nota(s) de R$ 100.00\n",hundred);

   printf("%d nota(s) de R$ 50.00\n",fifty);

   printf("%d nota(s) de R$ 20.00\n",twenty);

   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",coin1);

   printf("%d moeda(s) de R$ 0.50\n",coin2);

   printf("%d moeda(s) de R$ 0.25\n",coin3);

   printf("%d moeda(s) de R$ 0.10\n",coin4);

   printf("%d moeda(s) de R$ 0.05\n",coin5);

   printf("%.0f moeda(s) de R$ 0.01\n",coin6);

    return 0;

}