http://acm.hdu.edu.cn/showproblem.php?pid=3294

Girls‘ research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5711    Accepted Submission(s): 2117

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

Source

2010 “HDU-Sailormoon” Programming Contest

Recommend

lcy

题意：先转换，再输出最长回文串及下标。

思路：利用原数组与加工后的数组的数量关系得到下标。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#define INF  10000000
using namespace std;
char a[200009] , b[400009] , str[200009];
int p[400009];

int num;

int main()
{
    char c ;
    while(~scanf("%c%s" , &c , a))
    {
        int len = strlen(a);
        int q = c - ‘a‘;
        memset(str , 0 , sizeof(str));
        memset(p , 0 , sizeof(p));
        for(int i = 0 ; i < len ; i++)
        {
            if(a[i] - q < ‘a‘)
            {
                a[i] = a[i] + 26;
            }
            a[i] = a[i] - q ;
        }
        //printf("%s\n" , a);
        int l = 0 ;
        b[l++] = ‘$‘;
        b[l++] = ‘#‘;
        for(int i = 0 ; i < len ; i++)
        {
            b[l++] = a[i];
            b[l++] = ‘#‘;
        }
        int mx = - 1 , mid , ans = 0;
        for(int i = 1 ; i < l ; i++)
        {
            if(mx > i)
            {
                p[i] = min(p[2*mid-i] , mx - i);
            }
            else
            {
                p[i] = 1 ;
            }
            while(b[i-p[i]] == b[i+p[i]])
            {
                p[i]++;
            }
            if(mx < p[i]+i)
            {
                mid = i ;
                mx = p[i] + i;
            }
        }
        int index = 0  , indey = 0;
        for(int i = 0 ; i < l ; i++)
        {
            if(ans < p[i] - 1)
            {
                ans = p[i] - 1;
                index = i ;

            }
        }
        if(ans == 1)
        {
            printf("No solution!\n");
        }
        else
        {
            int r = (index + ans)/2 - 1;
            int l = r - ans + 1;
            printf("%d %d\n" , l , r);
            for(int i = l ; i <= r ; i++)
            {
                printf("%c" , a[i]);
            }
            printf("\n");

        }
        getchar();
    }

    return 0 ;
}

原文地址：https://www.cnblogs.com/nonames/p/11296274.html