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To Buy or Not to Buy (20)

[编程题]To Buy or Not to Buy (20)

热度指数：2337 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

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.

示例1

输入

ppRYYGrrYBR2258<br/>YrR8RrY

输出

Yes 8

算法知识视频讲解

Python

雨润田上

#Python3
color=list(input())
aim=list(input())
asci=[0]*128
lack=0
for item in color:
    asci[ord(item)]+=1
for item in aim:
    asci[ord(item)]-=1
    if asci[ord(item)]<0:
        lack+=1
        
if lack<=0:
    print('Yes'+' '+str(len(color)-len(aim)))
else:
    print('No'+' '+str(lack))

发表于 2021-01-24 23:41:02 回复(0)