PAT 1092 To Buy or Not to Buy python解法-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_36936510/article/details/85992881

本文解析了1092BuyorNottoBuy算法问题，该问题要求判断商店提供的珠串是否包含顾客所需的所有颜色珠子，并计算多余的珠子数量或缺失的珠子数量。通过将字符串转换为列表并进行比较，实现了算法的高效解决。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1092 To Buy or Not to Buy （20 分）
Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must check whether a string in the shop contains all the beads she needs. She now comes to you for help: if the answer is Yes, please tell her the number of extra beads she has to buy; or if the answer is No, please tell her the number of beads missing from the string.

For the sake of simplicity, let’s use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. For example, the 3rd string in Figure 1 is the one that Eva would like to make. Then the 1st string is okay since it contains all the necessary beads with 8 extra ones; yet the 2nd one is not since there is no black bead and one less red bead.

Figure 1

Input Specification:
Each input file contains one test case. Each case gives in two lines the strings of no more than 1000 beads which belong to the shop owner and Eva, respectively.

Output Specification:
For each test case, print your answer in one line. If the answer is Yes, then also output the number of extra beads Eva has to buy; or if the answer is No, then also output the number of beads missing from the string. There must be exactly 1 space between the answer and the number.

Sample Input 1:
ppRYYGrrYBR2258
YrR8RrY
Sample Output 1:
Yes 8
Sample Input 2:
ppRYYGrrYB225
YrR8RrY
Sample Output 2:
No 2

解题思路：虽然题目很长，但意思其实很简单，对于Input 来说，第一行是店家提供的字符串，第二行是你需要的字符串，我们做的就是看店家给的字符串是否包含你需要的字符串中所有字符，如果全部包含，则输出Yes以及需要额外购买多少个字符，否则输出No以及缺多少个字符。首先，将字符串转换为列表，循环扫描需要的字符串中的元素，判断是否在店家提供的字符串中，如果存在则删除，不存在则计数器加1，扫描完成后，如果计数器是0，说明全部包含，多出的字符个数就是店家提供的字符串长度减去需要的字符串长度，如果不等于0，则缺少的字符个数就是计数器的值。

str_in = input()
sub_str_in = input()
str_l = [i for i in str_in]
#str_l.sort() #排序之后耗时更短
sub_str_l = [i for i in sub_str_in]
#sub_str_l.sort()
count = 0
for i in sub_str_l:
    if i not in str_l:
        count += 1
    else:
        str_l.remove(i)

if count == 0:
    print('Yes',len(str_in)-len(sub_str_in))
else:
    print('No',count)