#include <iostream>
#include <cstdio>
using namespace std;
//小技巧;将浮点数转化为整数,本题乘以100即可
int main()
{
double m;
cin >> m;
int n = m * 100;
printf("NOTAS:\n");
printf("%d nota(s) de R$ 100.00\n", n / 10000); n %= 10000;
printf("%d nota(s) de R$ 50.00\n", n / 5000); n %= 5000;
printf("%d nota(s) de R$ 20.00\n", n / 2000); n %= 2000;
printf("%d nota(s) de R$ 10.00\n", n / 1000); n %= 1000;
printf("%d nota(s) de R$ 5.00\n", n / 500); n %= 500;
printf("%d nota(s) de R$ 2.00\n", n / 200); n %= 200;
printf("MOEDAS:\n");
printf("%d moeda(s) de R$ 1.00\n", n / 100); n %= 100;
printf("%d moeda(s) de R$ 0.50\n", n / 50); n %= 50;
printf("%d moeda(s) de R$ 0.25\n", n / 25); n %= 25;
printf("%d moeda(s) de R$ 0.10\n", n / 10); n %= 10;
printf("%d moeda(s) de R$ 0.05\n", n / 5); n %= 5;
printf("%d moeda(s) de R$ 0.01\n", n / 1); n %= 1;
}
#include<iostream>
using namespace std;
int main(){
double n;
cin>>n;
//钞票面额数组
double nota[6]={100,50,20,10,5,2};
//硬币面额数组
double modea[6]={1,0.5,0.25,0.1,0.05,0.01};
//钞票输出
cout<<"NOTAS:"<<endl;
for(int i=0;i<6;i++){
//num为当前面额为nota[i]的钞票所需张数
int num=int(n/nota[i]);
printf("%d nota(s) de R$ %.2lf\n",num,nota[i]);
n=n-num*nota[i];
}
//硬币输出
cout<<"MOEDAS:"<<endl;
for(int i=0;i<6;i++){
//num为当前面额为modea[i]的硬币所需个数,面额最小为10e-2
int num=int(n/modea[i]+10e-3);//通过加一个10e-3来防止精度问题
printf("%d moeda(s) de R$ %.2lf\n",num,modea[i]);
n=n-num*modea[i];
}
return 0;
}