HDU - 3294 Girls' research

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3523    Accepted Submission(s): 1344

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!

题意：给定一个key字母和字符串，后面字符串中的key实际上都对应为a字母，key+1对应b,以此类推

思路：裸manecher跑一遍，输出时按照密文转化关系输出

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define max_ 201000
#define inf 0x3f3f3f3f
#define mod 1000000007
using namespace std;
int n;
char ss[max_],s[max_*2],c;
int p[max_*2];
int main(int argc, char const *argv[]) {
    while(scanf(" %c",&c)!=EOF)
    {
        scanf(" %s",ss);
        s[0]='$';
        s[1]='#';
        int l=strlen(ss);
        for(int i=0;imx)
            {
                id=i;
                mx=i+p[i];
            }
        }
        int maxx=-1,k;
        for(int i=1;imaxx)
            {
                maxx=p[i];
                k=i;
            }
        }
        if(maxx==2)
        printf("No solution!\n" );
        else
        {
            printf("%d %d\n",(k-maxx)/2,(k+maxx-4)/2);
            for(int i=(k-maxx)/2;i<=(k+maxx-4)/2;i++)
            printf("%c",ss[i]-c+'a'>='a'?ss[i]-c+'a':'z'+ss[i]-c+1 );
            printf("\n" );
        }
    }
    return 0;
}