The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 D Carneginon
Carneginon was a chic bard. But when he was young, he was frivolous and had joined many gangs. Recently, Caneginon was to be crowned, because the king was shocked by his poems and decided to award him the gold medal lecturer. Therefore, Most of people in the Kingdom came to visit him.
However, as a medal lectirer, Carneginon must treat the visitors kindly, including elders and younger generations. In order to maintain order, every visitor received a license with a magic field engraved on it. And the magic field on the licence was made up of lowercase letters.
Carneginon had a unique licence, which could judge whether others are his older or younger. Now, we assume that the sequence on Carneginon's licence is TT and the sequence on visitors' licence is SS. For each visitor,
If the length of TT is longer than the length of SS, it's obviously that the visitor is younger. And if SS is a substring of TT, Carneginon would call the visitor
my child!. Otherwise, Carneginon would call the visitoroh, child!.If the length of TT is less than the length of SS, it's obviously that the visitor is elder. And if TT is a substring of SS, Carneginon would call the visitor
my teacher!. Otherwise, Carneginon would call the visitorsenior!.Of course, if the length of TT is equal to the length of SS, the visitor is Carneginon's peer. And if TT is equal to SS, it shows that the visitor entered through an improper way and Carneginon would shout
jntm!. Otherwise, Carneginon would call the visitorfriend!.
Now, you know the TT (Carneginon's licence), qq (the number of visitors) and each visitor's licence(S_iSi). Can you judge what Caneginon needs to say when he sees every visitor?
Input
The first line is a string TT, representing Carneginon's license.
The second line is and integer qq, which means the number of visitors.
Then mm lines, for each line, there is a string SS, denoting the visitor's license.
1 \leq |T| \leq 10^51≤∣T∣≤105, 1 \leq |S| \leq 10^51≤∣S∣≤105, 1 \leq q \leq 10001≤q≤1000
It is guaranteed that q \times (|S|+|T|) \leq 10^7q×(∣S∣+∣T∣)≤107.
Output
There are qq lines.
For each SS, output what Carneginon should say correctly.
样例输入复制
abcde
6
abcde
aaaaa
abcd
aaaa
abcdef
abccdefg
样例输出复制
jntm!
friend!
my child!
oh, child!
my teacher!
senior!
这个题简单KMP匹配,直接套模板,就ok,一A;
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
#define Swap(a,b) a^=b^=a^=b
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0)
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define INF 0x3f3f3f3f
#define maxn 100010
#define esp 1e-9
#define mp(a,b) make_pair(a,b)
using namespace std;
typedef long long ll;
//-----------------------*******----------------------------//
int nextt[100005];
void Getnextt(string &p,int nextt[])
{
int pLen = p.size();
nextt[0] = -1;
int k = -1;
int j = 0;
while (j < pLen - 1)
{
//p[k]表示前缀,p[j]表示后缀
if (k == -1 || p[j] == p[k])
{
++k;
++j;
nextt[j] = k;
}
else
{
k = nextt[k];
}
}
}
bool KmpSearch(string &s, string &p)
{
int i = 0;
int j = 0;
int sLen = s.size();
int pLen = p.size();
while (i < sLen && j < pLen)
{
//①如果j = -1,或者当前字符匹配成功(即S[i] == P[j]),都令i++,j++
if (j == -1 || s[i] == p[j])
{
i++;
j++;
}
else
{
//②如果j != -1,且当前字符匹配失败(即S[i] != P[j]),则令 i 不变,j = nextt[j]
//nextt[j]即为j所对应的nextt值
j = nextt[j];
}
}
if (j == pLen)
return true;
else
return false;
}
int main()
{
string s,w;
int n;
cin>>s>>n;
int len=s.size();
while(n--)
{ memset(nextt,0,sizeof(nextt));
cin>>w;
int x=w.size();
// cout<<x<<' '<<len<<endl;
if(x==len)
{
// cout<<-1<<endl;
if(w==s) puts("jntm!");
else puts("friend!");
}
else if(x<len)
{ Getnextt(w,nextt);
if(KmpSearch(s, w)) //w s
puts("my child!");
else
puts("oh, child!");
}
else
{
Getnextt(s,nextt);
if(KmpSearch(w, s))
puts("my teacher!");
else puts("senior!");
}
}
}
The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 D Carneginon的更多相关文章
- The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 K题 center
You are given a point set with nn points on the 2D-plane, your task is to find the smallest number o ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 XKC's basketball team
XKC , the captain of the basketball team , is directing a train of nn team members. He makes all mem ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 C Buy Watermelon
The hot summer came so quickly that Xiaoming and Xiaohong decided to buy a big and sweet watermelon. ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 B so easy
题目链接:https://nanti.jisuanke.com/t/41384 这题暴力能过,我用的是并查集的思想,这个题的数据是为暴力设置的,所以暴力挺快的,但是当他转移的点多了之后,我觉得还是我这 ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 A Who is better?
A After Asgard was destroyed, tanker brought his soldiers to earth, and at the same time took on the ...
- 计蒜客 41391.query-二维偏序+树状数组(预处理出来满足情况的gcd) (The Preliminary Contest for ICPC Asia Xuzhou 2019 I.) 2019年徐州网络赛)
query Given a permutation pp of length nn, you are asked to answer mm queries, each query can be rep ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 E XKC's basketball team [单调栈上二分]
也许更好的阅读体验 \(\mathcal{Description}\) 给n个数,与一个数m,求\(a_i\)右边最后一个至少比\(a_i\)大\(m\)的数与这个数之间有多少个数 \(2\leq n ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019
A:Who is better? 题目链接:https://nanti.jisuanke.com/t/41383 题意: 类似于有N个石子,先手第一次不能拿完,每次后手只能拿 1 到 前一次拿的数量* ...
- The Preliminary Contest for ICPC Asia Xuzhou 2019 E. XKC's basketball team
题目链接:https://nanti.jisuanke.com/t/41387 思路:我们需要从后往前维护一个递增的序列. 因为:我们要的是wi + m <= wj,j要取最大,即离i最远的那个 ...
随机推荐
- Eclipse(Eclipse for android)代码自动提示设置
java代码和xml资源代码在有的eclipse中可以自己提示,但有的并不支持这个功能,还得我们人为去调整,主要原因是因为你们下载的elipse的渠道不同,获得的版本有的官方原版,有的是个人备份版等等 ...
- bootstraptable 必备知识点
1.如何动态刷新表中数据? (1).无参刷新: $("#table").bootstrapTable('refresh'); (2).带参刷新: var opt = { url: ...
- 28.3 api--date 日期 (日期获取、格式化)
/* * Date: 表示特定的瞬间,精确到毫秒,他可以通过方法来设定自己所表示的时间,可以表示任意的时间 * System.currentTimeMillis():返回的是当前系统时间,1970-1 ...
- CH5E07 划分大理石(背包dp+二进制拆分)
传送门 大意: 有价值分别为1..6的大理石各a[1..6]块,现要将它们分成两部分,使得两部分价值之和相等,问是否可以实现.其中大理石的总数不超过20000. 解题思路: 妥妥的多重背包+二 ...
- canal使用记录
canal是阿里巴巴的来源项目.我们可以通过配置binlog实现数据库监控,得到数据库表或者数据的更新信息.参考我的文档前先去官网看下,可能已经支持更高版本的MySQL了 1. 查看官方开源项目 ht ...
- vue-cli3 按需引入 element-ui 报错
报错信息: Cannot find module 'babel-preset-es2015' from .... 解决办法: 安装最新的 Babel 编译插件:@babel/preset-env 修改 ...
- 智能指针 shared_ptr
1.不支持数组 2.c++11支持make_shared,分配一次内存,构造函数为private和proteced时不能调用. 3.new初始化分配两次内存,一.分配数据块内存,二.分配控制块内存
- 如何使用Three.js加载obj和mtl文件
OBJ和MTL是3D模型的几何模型文件和材料文件. 在最新的three.js版本(r78)中,以前的OBJMTLLoader类已废弃. 现在要加载OBJ和MTL文件,需要结合OBJLoader和MTL ...
- 你只要5行代码,拥有你的个性二维码,用Python生成动态二维码
如果想了解更多关于python的应用,可以私信我,或者点击下方链接自行获取,里面到资料都是免费的(http://t.cn/A6Zvjdun) 二维码满天飞,但是有没有想过Python也能制作出专属于自 ...
- 食物链 POJ - 1182 (并查集的两种写法)
这是一个非常经典的带权并查集,有两种写法. 1 边权并查集 规定一下,当x和y这条边的权值为0时,表示x和y是同类,当为1时,表示x吃y,当为2时,表示x被y吃. 一共有三种状态,如图,当A吃B,B吃 ...