PAT甲级(Advanced Level)真题--1046 Sharing

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本文解析了PAT甲级(Advanced Level)的一道真题——1046 Sharing，介绍了如何使用集合去重来找到两个单词共享后缀的起始位置。通过分析输入输出描述，给出了一段Python代码示例。

PAT甲级(Advanced Level)真题–1046 Sharing

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To store English words, one method is to use linked lists and store a
word letter by letter. To save some space, we may let the words share
the same sublist if they share the same suffix. For example,
“loading” and “being” are stored as showed in Figure
在这里插入图片描述

You are supposed to find the starting position of the common
suffix (e.g. the position of “i” in Figure 1).

输入描述:
Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (<= 105), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is the letter contained by this node which is an English letter chosen from {a-z, A-Z}, and Next is the position of the next node.

输出描述:
For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output “-1” instead.
对于每种情况，只需输出公共后缀的5位起始位置。如果这两个单词没有共同的后缀，则输出“-1”。

输入例子:

11111 22222 9
67890 i 00002
00010 a 12345
00003 g -1
12345 D 67890
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 67890
00001 o 00010

输出例子:

题目很简单,使用集合去重就可以了。

Python代码如下：

x=input().strip().split(' ')
temp=set()
flag=0
for i in range(0,int(x[2])):
    y=input().strip().split(' ')
    if y[2] in temp:
        flag=y[2]
    else:
        temp.add(y[2])
if flag==0:
               print("-1")
else:
               print(flag)