Python贪心算法实战：找零、最大子数组与加油站问题详解

# -*- coding:utf-8 -*-
def main():
  d = [0.01,0.02,0.05,0.1,0.2,0.5,1.0] # 存储每种硬币面值
  d_num = [] # 存储每种硬币的数量
  s = 0
  # 拥有的零钱总和
  temp = raw_input('请输入每种零钱的数量：')
  d_num0 = temp.split(" ")
  for i in range(0, len(d_num0)):
    d_num.append(int(d_num0[i]))
    s += d[i] * d_num[i] # 计算出收银员拥有多少钱
  sum = float(raw_input("请输入需要找的零钱:"))
  if sum > s:
    # 当输入的总金额比收银员的总金额多时，无法进行找零
    print("数据有错")
    return 0
  s = s - sum
  # 要想用的钱币数量最少，那么需要利用所有面值大的钱币，因此从数组的面值大的元素开始遍历
  i = 6
  while i >= 0: 
    if sum >= d[i]:
      n = int(sum / d[i])
      if n >= d_num[i]:
        n = d_num[i] # 更新n
      sum -= n * d[i] # 贪心的关键步骤，令sum动态的改变，
      print("用了%d个%f元硬币"%(n, d[i]))
    i -= 1
if __name__ == "__main__":
  main()

# -*- coding:utf-8 -*-
def main():
  s = [12,-4,32,-36,12,6,-6]
  print("定义的数组为：",s)
  s_max, s_sum = 0, 0
  for i in range(len(s)):
    s_sum += s[i]
    if s_sum >= s_max:
      s_max = s_sum # 不断更新迭代s_max的值，尽可能的令其最大
    elif s_sum < 0:
      s_sum = 0
  print("最大子数组和为：",s_max)
if __name__ == "__main__":
  main()

# 设汽车加满油后可行驶n公里，且旅途中有k个加油站
def greedy():
  n = 100
  k = 5
  d = [50,80,39,60,40,32]
  # 表示加油站之间的距离
  num = 0
  # 表示加油次数
  for i in range(k):
    if d[i] > n:
      print('no solution')
      # 如果距离中得到任何一个数值大于n 则无法计算
      return 
  i, s = 0, 0
  # 利用s进行迭代
  while i <= k:
    s += d[i]
    if s >= n:
      # 当局部和大于n时则局部和更新为当前距离
      s = d[i]
      # 贪心意在令每一次加满油之后跑尽可能多的距离
      num += 1
    i += 1
  print(num)
if __name__ == '__main__':
  greedy()