kuangbin J
来源:互联网 发布:数据库客户端如何使用 编辑:程序博客网 时间:2024/05/20 04:31
Problem Description
Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.
Input
Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.
Output
Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.
Sample Input
clinton
homer
riemann
marjorie
Sample Output
0
rie 3
题解:
next数组的应用。
连接两个字符串求next[s1+s2];
sprintf可以把数据打印到字符串中
但是注意next[s1+s2]不得大于strlen(s1)和strlen(s2)
这里WA了一次。
如
abcabc
abc
next[len]=6
输出3
abc
abcabc
next[len]=6
数出3
代码:
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 50000 + 10;typedef long long LL;char str[MAXN+MAXN];LL len;LL nt[MAXN+MAXN];void getNext(){ nt[0] = -1; LL i = 0, j = -1; while (i <= len) { if (j == -1 || str[i] == str[j]) { nt[++i] = ++j; } else { j = nt[j]; } }}int main(){ char s1[MAXN],s2[MAXN]; while(gets(s1),gets(s2)) { sprintf(str,"%s%s",s1,s2); len = strlen(str); getNext(); LL ans = nt[len]; if(ans>strlen(s1)) ans=strlen(s1); if(ans>strlen(s2)) ans=strlen(s2); if(ans==0) { cout<<0<<endl; } else { for(int i=0;i<ans;i++) printf("%c",str[i]); printf(" %lld\n",ans); } } return 0;}
- kuangbin J
- kuangbin 简单搜索 J 双bfs
- [kuangbin带你飞]专题一 简单搜索-J - Fire!
- [kuangbin带你飞]专题一 简单搜索 J
- [kuangbin带你飞]专题一 简单搜索J - Fire!(UVA 11624)
- [kuangbin带你飞]专题一 简单搜索 J - Fire! UVA 11624
- [kuangbin带你飞]专题一 简单搜索 J UVA 11624
- [kuangbin带你飞]专题七 线段树 J HDU 3974
- [kuangbin带你飞]专题四 最短路练习 J POJ 1511
- [kuangbin带你飞]专题五 并查集 J POJ 2492
- [kuangbin带你飞]专题六 最小生成树 J POJ3026
- [kuangbin带你飞]专题十二 基础DP1 J HDU 1160
- [kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher J HDU 2594
- [kuangbin带你飞]专题四 最短路练习H,I,J
- [kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher J (kmp扩展)
- J
- j
- j
- mysql table相关命令
- hdu6200mustedge mustedge mustedge
- 模型分类评价
- SpringBoot配置log4j2的JdbcAppender日志写入数据库,可定义哪些日志写入
- Docker for Windows 里的Shared Drives 设置不生效
- kuangbin J
- Java内部类详解
- 上拉电阻与下拉电阻
- 【复赛模拟试题】求和 分治+二分快速幂
- Linux中exec命令相关
- 配置好apache SSL之后 浏览器https访问未响应的解决方法
- Andorid事件分发源码解析
- 图论的题目
- Linux网络编程