PAT甲级(Advanced Level)真题-- 1062 To Buy or Not to Buy

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提交：1220
通过率：52%

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 string “YrR8RrY” is the one that Eva would like
to make. Then the string “ppRYYGrrYBR2258” is okay since it
contains

all the necessary beads with 8 extra ones; yet the string
“ppRYYGrrYB225” is not since there is no black bead and one
less red bead.

输入描述:
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.

输出描述:
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.

输入例子:
ppRYYGrrYBR2258
YrR8RrY

输出例子:
Yes 8

对于每个测试用例，请将答案打印在一行中。如果答案是“是”，那么还要输出Eva必须购买的额外珠子的数量; 或者
如果答案是“否”，则还输出字符串中缺少的珠子数量。

这道题很简单，python解法如下：

x=input().strip()
y=input().strip()
flag=0
a=set()
count=0
for i in y:a.add(i)
for i in a:if x.count(i)<y.count(i):flag=1count+=y.count(i)-x.count(i)
if flag==1:print("No",count)
else:print("Yes",len(x)-len(y))

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PAT甲级(Advanced Level)真题-- 1062 To Buy or Not to Buy

PAT甲级(Advanced Level)真题-- 1062 To Buy or Not to Buy

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