1032 Sharing (25)(25 分) 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 "

1032 Sharing (25 分)   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 store

1032 Sharing (25分) 题目 思路 定义map存储所有的<地址1,地址2> 第一set存放单词1的所有地址(通过查找map) 通过单词二的首地址,结合map,然后在set中查找是否存在,如果存在就是所求的地址,没有就是-1 注意点 无 代码 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #in

题意: 输入两个单词的起始地址和一个正整数N(<=1e5),然后输入N行数据,每行包括一个五位数的字母地址,字母和下一个字母的地址.输出这两个单词的公共后缀首字母的地址,若无公共后缀则输出-1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string a1,a2; int n; string s,nex; char x; map<strin

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

题目 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 Figu

1032. Sharing (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 the

题目 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 sufix. For example, "loading" and "being" are stored

题目如下: 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 st

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 a

#include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> using namespace std; /* 链表题 水 */ int n; struct Word{ int addr; char ch; ; }word[]; ]; int main() { int first1,first2; int adr,nxt

problem 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

简单题,不过数据中好像存在有环的链表...... #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<vector> using namespace std; +; struct Node { int id;

1032 Sharing (25)(25 分) 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 "

1032. Sharing (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 the

1114 Family Property (25 分) This time, you are supposed to help us collect the data for family-owned property. Given each person's family members, and the estate(房产)info under his/her own name, we need to know the size of each family, and the average

1044 Shopping in Mars (25 分) Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and som

题目地址 https://pta.patest.cn/pta/test/15/exam/4/question/723 5-15 QQ帐户的申请与登陆   (25分) 实现QQ新帐户申请和老帐户登陆的简化版功能.最大挑战是:据说现在的QQ号码已经有10位数了. 输入格式: 输入首先给出一个正整数NN(\le 10^5≤10​5​​),随后给出NN行指令.每行指令的格式为:“命令符(空格)QQ号码(空格)密码”.其中命令符为“N”(代表New)时表示要新申请一个QQ号,后面是新帐户的号码和密码:命令

B1020 月饼(25 分) 月饼是中国人在中秋佳节时吃的一种传统食品,不同地区有许多不同风味的月饼.现给定所有种类月饼的库存量.总售价.以及市场的最大需求量,请你计算可以获得的最大收益是多少. 注意:销售时允许取出一部分库存.样例给出的情形是这样的:假如我们有 3 种月饼,其库存量分别为 18.15.10 万吨,总售价分别为 75.72.45 亿元.如果市场的最大需求量只有 20 万吨,那么我们最大收益策略应该是卖出全部 15 万吨第 2 种月饼.以及 5 万吨第 3 种月饼,获得 72 +

A1009 Product of Polynomials (25)(25 分) This time, you are supposed to find A*B where A and B are two polynomials. Input Specification: Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a po