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Copy path最少换钱问题.py
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最少换钱问题.py
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time : 2019/9/6 20:16
# @Author : SeniorZhu1994
# @Email : SeniorZhu1994@
# @Site :
# @File : 最少换钱问题.py
# @Software: PyCharm
# @Describe: https://blog.csdn.net/qq_34342154/article/details/77104520
# !/bin/python
# -*- coding: utf8 -*-
import sys
import os
import re
def minCoins2(arr, aim):
if arr == None or len(arr) == 0 or aim < 0:
return -1
row = len(arr)
dp = [sys.maxsize for i in range(aim+1)]
dp[0] = 0
for i in range(1, aim+1):
if i % arr[0] == 0:
dp[i] = i // arr[0]
for i in range(1, row):
dp[0] = 0
for j in range(1, aim+1):
left = sys.maxsize
if j - arr[i] >= 0 and dp[j-arr[i]] != sys.maxsize:
left = dp[j-arr[i]] + 1
dp[j] = min(left, dp[j])
return dp[aim] if dp[aim] != sys.maxsize else -1
# 请完成下面这个函数,实现题目要求的功能
# 当然,你也可以不按照下面这个模板来作答,完全按照自己的想法来 ^-^
# ******************************开始写代码******************************
def solution(prices, budget):
if prices == None or len(prices) == 0 or budget < 0:
return -1
row = len(prices)
dp = [[sys.maxsize for i in range(budget + 1)] for j in range(row)]
for i in range(row):
dp[i][0] = 0
for j in range(1, budget + 1):
if j % prices[0] == 0:
dp[0][j] = j // prices[0]
for i in range(1, row):
for j in range(1, budget + 1):
left = sys.maxsize
if j - prices[i] >= 0 and dp[i][j - prices[i]] != sys.maxsize:
left = dp[i][j - prices[i]] + 1
dp[i][j] = min(left, dp[i - 1][j])
return dp[row - 1][budget] if dp[row - 1][budget] != sys.maxsize else -1
def coins5(arr, aim):
if arr == None or len(arr) == 0 or aim < 0:
return 0
dp = [0 for i in range(aim+1)]
for i in range(aim//arr[0]+1):
dp[arr[0]*i] = 1
for i in range(len(arr)):
for j in range(1, aim+1):
dp[j] += dp[j-arr[i]] if j-arr[i] >= 0 else 0
return dp[aim]
# ******************************结束写代码******************************
_prices_cnt = 0
_prices_cnt = int(input())
_prices_i = 0
_prices = []
while _prices_i < _prices_cnt:
_prices_item = int(input())
_prices.append(_prices_item)
_prices_i += 1
_budget = int(input())
res = solution(_prices, _budget)
print(str(res) + "\n")