标签：only who 生成 you size something Once ant 开始
Please tell Cyy whether this circular string is symmetrical.
The input file consists of multiple test cases.
For each test case, there’s only a string in a line to represent the circular string. The string only consists of lowercase characters. ( 1 <= N <= 65536 )
It’s guaranteed that the sum of N is not larger than 1000000.
For each test case, output “Yes” in a line if the circular string is symmetrical, output “No” otherwise.
aaaaaa abcde ababab aaaaba aabbaa
Yes No No No Yes
一句话题意：循环字符串能否断开形成回文串
循环的东西，很明显要直接复制一遍
回文串，很明显manacher
如果原串长度len为奇数，则找到的回文数长度m必须>=len且m为奇数
如果原串长度len为偶数，则找到的回文数长度m必须>=len且m为偶数
一开始总是超时，一直以为是判断输入结束不对，后来才发现是用了string，每次都用string生成！#a#b..,很耗时！
1 #include<iostream> 2 #include<stdio.h> 3 #include<math.h> 4 #include<stdlib.h> 5 #include<string.h> 6 #include<algorithm> 7 using namespace std; 8 9 int p[300000]; 10 bool manacher(char* s,int len){ 11 int length=0; 12 int maxRight=1; 13 int mid=1; 14 int l=strlen(s); 15 for(int i=1;i<l;++i) 16 { 17 if(i<maxRight) 18 p[i]=min(p[2*mid-i],maxRight-i); 19 else p[i]=1; 20 21 while(s[i-p[i]]==s[i+p[i]]) p[i]++; 22 23 if(p[i]+i>maxRight) 24 { 25 maxRight=p[i]+i; 26 mid=i; 27 } 28 length=p[i]-1; 29 if(len%2==0) 30 { 31 if(!(length<len||length%2==1)) 32 return true; 33 } 34 if(len%2==1) 35 { 36 if(!(length<len||length%2==0)) 37 return true; 38 } 39 } 40 return false; 41 42 } 43 44 int main() 45 { 46 int len; 47 char s[300000]; 48 49 while(~scanf("%s",&s)) 50 { 51 len=strlen(s); 52 for(int i=len;i>=0;--i) 53 s[i+len]=s[i]; 54 // cout<<s<<endl; 55 for(int i=2*len;i>=0;--i) 56 { 57 s[2*i+2]=s[i]; 58 s[2*i+1]=‘#‘; 59 } 60 61 s[0]=‘!‘; 62 // cout<<s<<endl; 63 64 if(manacher(s,len)==true) 65 cout<<"Yes"<<endl; 66 else cout<<"No"<<endl; 67 } 68 69 }
标签：only who 生成 you size something Once ant 开始
原文地址：https://www.cnblogs.com/noip/p/9570168.html
