用到了贪心的思想，所有这在语法题里算中等难度了！

// Greedy Solution
#include <iostream>

using namespace std;
int n;

int denominations[] {1, 2, 5, 10, 20, 50, 100};
int counts[7];

int main(void) {
  scanf("%d", &n);
  const int x = n;
  for (int i = 6; i >= 0; --i) {
    while (n >= denominations[i]) {
      n -= denominations[i];
      ++counts[i];
    }
  }

  printf("%d\n", x);
  for (int i = 6; i >= 0; --i)
    printf("%d nota(s) de R$ %d,00\n", counts[i], denominations[i]);

  return 0;
}