Simpsons’ Hidden Talents HDU
来源:互联网 发布:mac禁止程序自动启动 编辑:程序博客网 时间:2024/06/03 20:13
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.
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.
The lengths of s1 and s2 will be at most 50000.
clintonhomerriemannmarjorie
0rie 3
思路:两个字符串合并成一个,然后kmp,注意,这时候算出来的应该是串长度应该是小于len1与len2的。
#include<stdio.h>#include<string.h>using namespace std;const int M=2e5+10;char str1[M],str2[M];int next[M];void getnext(int len){ next[0]=-1; for(int i=1; i<len; i++) { int j=next[i-1]; while(j>=0&&str1[i]!=str1[j+1]) j=next[j]; if(str1[i]==str1[j+1]) next[i]=j+1; else next[i]=-1; }}int main(){ while(scanf("%s%s",str1,str2)!=EOF) { int olen=strlen(str1); int olen2=strlen(str2); strcat(str1,str2); int len=strlen(str1); getnext(len); int j=next[len-1]; while(j>=olen||j>=olen2) j=next[j]; if(j==-1) printf("%d\n",0); else { for(int i=0; i<=j; i++) printf("%c",str1[i]); printf(" %d\n",j+1); } }}
阅读全文
0 0
- HDU Simpsons’ Hidden Talents
- Simpsons’ Hidden Talents HDU
- Simpsons’ Hidden Talents HDU
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- HDU 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- HDU 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- HDU - 2594 Simpsons’ Hidden Talents
- HDU 2594 Simpsons’ Hidden Talents
- HDU 2594 Simpsons’ Hidden Talents
- hdu 2594 Simpsons’ Hidden Talents
- webpack 配置学习笔记
- 提高 Java 代码性能的各种技巧
- codeforces839C Journey(DFS)
- 数据结构实验之二叉树四:还原二叉树
- 深拷贝、浅拷贝、数据类型
- Simpsons’ Hidden Talents HDU
- pycharm 安装
- java集合
- tomcat启动startup.bat闪退
- 使用Python进行AES加密和解密
- MVP模式初探
- docker 基本使用
- 561. Array Partition I
- 动态规划——背包问题