编写HackerRank消息解码程序时遭遇EOFError问题求助

问题描述 #

编写HackerRank解码程序时遇到EOFError，程序功能为接收映射字典msg_dict、消息数量整数no_of_messages、待解码消息列表，返回无空格的解码消息列表。错误信息如下：

Traceback (most recent call last):
  File "/tmp/submission/20221104/11/52/hackerrank-cc4efe99dc145dc161f1a763dbcdf2e1/code/Solution.py", line 56, in <module>
    msg_dict_count = int(input().strip())
EOFError: EOF when reading a line

预期运行效果 #

示例输入0：

9
r 5
l 7
w 3
h 1
d 0
u 4
o 9
y 2
e 6
2
d h y u y r w u u r
y h w e u h e h d d

示例输出0：

hello
world

当前代码 #

#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'decode_messages' function below.
#
# The function is expected to return a LIST_OF_STRINGS - the list of decoded messages.
# The function accepts the following parameters:
#  1. DICTIONARY msg_dict - the dictionary to be used to decode each character
#  2. INTEGER no_of_messages - the number of messages in the list to be decoded
#  3. LIST_OF_STRINGS messages - the list of messages with each character (in each message) separated by space
#

 

# Write your code here
def decode_messages(msg_dict,no_of_messages,messages):
    decodeList=[]
    for i in range(no_of_messages):
        decodeList.append([])
        for j in range(0,len(messages[i]),2):
            if messages[i][j+1]!='':
                decodeList[i].append(msg_dict[messages[i][j]]+msg_dict[messages[i][j+1]]) #add consecutive characters of messages
    return decodeList
n=int(input())
msg_dict={} #storing pair consisting of letter and key
messages=[]
for i in range(n):
    a,b=input().split()
    msg_dict[a]=int(b)
no_of_messages=int(input())
for i in range(no_of_messages):
    messages.append(input().split())
decoded_list=decode_messages(msg_dict,no_of_messages,messages)
keys=list(msg_dict.keys())
values=list(msg_dict.values())
decodedMsgs=""
for i in decoded_list:
    for j in i:
        decodedMsgs=decodedMsgs+keys[values.index(j)] #get key corresponding to 
    print(decodedMsgs)
    decodedMsgs=""
    
   

if __name__ == '__main__':
   fptr = open(os.environ['OUTPUT_PATH'], 'w')

   msg_dict_count = int(input().strip())

   msg_dict = {}

   for _ in range(msg_dict_count):
       key_val = input().rstrip().split()
       msg_dict[key_val[0]] = key_val[1]

   messages_count = int(input().strip())

   messages = []

   for _ in range(messages_count):
       messages_item = input()
       messages.append(messages_item)

   decoded_msgs = decode_messages(msg_dict, messages_count, messages)

   fptr.write('\n'.join(decoded_msgs))
   fptr.write('\n')

   fptr.close()

 
   

错误原因与解决方法 #

核心问题 #

你的代码存在重复读取输入的问题：

1. 在if __name__ == '__main__':块之外，你已经写了一整套读取输入、调用函数、打印结果的代码；
2. HackerRank的评判系统会执行if __name__ == '__main__':块中的代码，这部分会再次尝试读取输入，但此时输入已被前面的代码读完，导致后续input()调用遇到EOF（文件结束符），触发EOFError。

同时，decode_messages函数逻辑错误，无法生成预期的解码结果，需要一并修正。

分步解决 #

1. 删除冗余的输入读取代码
   删掉if __name__ == '__main__':块之外的所有测试代码（从def decode_messages之后到if __name__之前的部分），只保留函数定义和if __name__块的代码。

2. 修正decode_messages函数逻辑
   原函数逻辑完全错误，需重新实现：

   • 反转msg_dict，构建数字字符串:字符的映射，方便通过拼接后的数字查找目标字符；
   • 对每条消息按空格分割成字符列表，每两个字符为一组，拼接对应数字字符串；
   • 通过反转后的字典找到对应字符，拼接成完整字符串后加入结果列表。

修正后的函数：

def decode_messages(msg_dict, no_of_messages, messages):
    # 反转字典，构建数字到字符的映射
    reversed_dict = {v: k for k, v in msg_dict.items()}
    decoded = []
    for msg in messages:
        chars = msg.split()
        result = []
        # 每两个字符一组处理
        for i in range(0, len(chars), 2):
            # 拼接两个数字字符串
            num_str = msg_dict[chars[i]] + msg_dict[chars[i+1]]
            # 找到对应的字符并添加到结果
            result.append(reversed_dict[num_str])
        # 拼接成完整字符串加入列表
        decoded.append(''.join(result))
    return decoded

3. 保持字典值类型一致
   在if __name__块中，读取的字典值是字符串类型，无需转换为整数，确保反转字典时类型匹配。

最终完整代码 #

#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'decode_messages' function below.
#
# The function is expected to return a LIST_OF_STRINGS - the list of decoded messages.
# The function accepts the following parameters:
#  1. DICTIONARY msg_dict - the dictionary to be used to decode each character
#  2. INTEGER no_of_messages - the number of messages in the list to be decoded
#  3. LIST_OF_STRINGS messages - the list of messages with each character (in each message) separated by space
#

def decode_messages(msg_dict, no_of_messages, messages):
    reversed_dict = {v: k for k, v in msg_dict.items()}
    decoded = []
    for msg in messages:
        chars = msg.split()
        result = []
        for i in range(0, len(chars), 2):
            num_str = msg_dict[chars[i]] + msg_dict[chars[i+1]]
            result.append(reversed_dict[num_str])
        decoded.append(''.join(result))
    return decoded

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    msg_dict_count = int(input().strip())

    msg_dict = {}

    for _ in range(msg_dict_count):
        key_val = input().rstrip().split()
        msg_dict[key_val[0]] = key_val[1]

    messages_count = int(input().strip())

    messages = []

    for _ in range(messages_count):
        messages_item = input()
        messages.append(messages_item)

    decoded_msgs = decode_messages(msg_dict, messages_count, messages)

    fptr.write('\n'.join(decoded_msgs))
    fptr.write('\n')

    fptr.close()

内容的提问来源于stack exchange，提问作者user20413640