Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 2176    Accepted Submission(s): 831 Problem Description One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps: First step: girls will write a long string (only contains lower case) on the paper. For example, "abcde", but 'a' inside is not the real 'a', that means if we define the 'b' is the real 'a', then we can infer that 'c' is the real 'b', 'd' is the real 'c' ……, 'a' is the real 'z'. According to this, string "abcde" changes to "bcdef". Second step: girls will find out the longest palindromic string in the given string, the length of palindromic string must be equal or more than 2.   Input Input contains multiple cases. Each case contains two parts, a character and a string, they are separated by one space, the character representing the real 'a' is and the length of the string will not exceed 200000.All input must be lowercase. If the length of string is len, it is marked from 0 to len-1.   Output Please execute the operation following the two steps. If you find one, output the start position and end position of palindromic string in a line, next line output the real palindromic string, or output "No solution!". If there are several answers available, please choose the string which first appears.   Sample Input b babd a abcd   Sample Output 0 2 aza No solution!   Author wangjing1111  

裸的马拉车问题。可以在O（n）时间内求解最长回文子串。

具体的实现原理可以看这篇博客http://www.cnblogs.com/biyeymyhjob/archive/2012/10/04/2711527.html

#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <string> using namespace std; const int MAXN=500010; char s[MAXN]; void init(int n){ int len =strlen(s); for(int i=0;i<len;i++){ int x=s[i]-'a'; x=(x+n+26)&; s[i]='a'+x; } } int p[MAXN]; void findBMstr(char* str,int n) { //int n=strlen(str); memset(p, 0, sizeof(p)); int mx = 0, id = 0; for(int i = 1; i <= n; i++){ if(mx > i){ p[i] = (p[2*id - i] < (mx - i) ? p[2*id - i] : (mx - i)); } else{ p[i] = 1; } while(str[i - p[i]] == str[i + p[i]]) p[i]++; if(i + p[i] > mx){ mx = i + p[i]; id = i; } } int max = 0, ii; for(int i = 1; i < n; i++){ if(p[i] > max){ ii = i; max = p[i]; } } max--; int start = ii - max ; int end = ii + max; if(max==1){ puts("No solution!"); return; } printf("%d %d\n",(start-1)/2,(end-1)/2-1); //cout<<(start-1)/2-1<<" "<<(end-1)/2-1<<endl; for(int i = start; i <= end; i++){ if(str[i] != '#'){ printf("%c",str[i]); } } printf("\n"); } char str0[MAXN]; int main(int argc, char *argv[]) { //ios::sync_with_stdio ( false ); char temp[2]; while(scanf("%s%s",temp,s)!=EOF){ int n=temp[0]-'a'; init(-n); str0[0]='$',str0[1]='#'; int k=2; int len=strlen(s); for(int i = 0; i < len; i++){ str0[k++]=s[i]; str0[k++]='#'; } //printf("%s\n",str0); findBMstr(str0,k); } }