Simpsons’ Hidden TalentsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12391 Accepted Submission(s): 4296 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. 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. Sample Input clinton homer riemann marjorie Sample Output 0 rie 3 Source HDU 2010-05 Programming Contest Recommend lcy | We have carefully selected several similar problems for you: 2595 2596 2597 2598 2599 |
题解:
为了练exkmp,这题用了exkmp来打。
代码(模板):
#include<bits/stdc++.h>
using namespace std;
int ans,nexr[200001],ex[200001];
void getnexr(char s[]){
int i=0,j,po,len=strlen(s);
nexr[0]=len;
while(s[i]==s[i+1]&&i+1<len)i++;
nexr[1]=i;
po=1;
for(i=2;i<len;i++){
if(nexr[i-po]+i<nexr[po]+po)nexr[i]=nexr[i-po];
else{
j=nexr[po]+po-i;
if(j<0)j=0;
while(i+j<len&&s[j]==s[i+j])j++;
nexr[i]=j;
po=i;
}
}
}
void exkmp(char s1[],char s2[]){
int i=0,j,po=0,len=strlen(s1),l2=strlen(s2);
while(s1[i]==s2[i]&&i<l2&&i<len)i++;
ex[0]=i;
for(i=1;i<len;i++){
if(nexr[i-po]+i<ex[po]+po)ex[i]=nexr[i-po];
else{
j=ex[po]+po-i;
if(j<0)j=0;
while(i+1<len&&s1[j+i]==s2[j])j++;
ex[i]=j;
po=i;
}
}
}
int main(){
int i;
char s1[200001],s2[200001];
while(~scanf("%s%s",s1,s2)){
getnexr(s1);
exkmp(s2,s1);
for(i=0;i<strlen(s2);i++)ans=max(ans,ex[i]);
for(i=0;i<ans;i++)printf("%c",s1[i]);
if(ans)printf(" %d\n",ans);
else printf("%d\n",ans);
ans=0;
}
}