Girls' research
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4401 Accepted Submission(s): 1642
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：通过第一个字符与a的关系翻译字符串，输出最长回文串和首尾下标，不存在则输出No solution!
分析：用manachar求出最长回文串中心和半径，因为变换后的串各字符下标改变了，所以输出原首尾下标要公式倒推
输出字符时要跳过插入的符号
0 1 2 3 4 5 6 7 8 9
$ * b * a * b * d *
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
char s[400001];
int p[400001];//半径
int main()
{
char c;
while (~scanf("%s%s", &c, s))
{
int len = strlen(s);
for (int i = len; i >= 0; i--)
s[2 * i + 2] = s[i],
s[2 * i + 1] = '*';
s[0] = '$';
int r = 0, mid = 0;//最大半径及对应回文串中心
int id = 0, mx = 0;//半径延申到最右字符时的点和当时遍历到的最右字符的下标
for (int i = 1; i < len * 2 + 1; i++)
{
if ( i < mx)//小于最右点
p[i] = min(p[2 * id - i], mx - i);//右端点不超出mx时，初始半径与关于id对称的点相等
else
p[i] = 1;//等于零下面也同样会变成1
while (s[i + p[i]] == s[i - p[i]])
p[i]++;//p[i]-1才是半径
if (i + p[i] > mx)
id = i, mx = i + p[i];
if (r < p[i] - 1)
r = p[i] - 1, mid = i;
}
if (r < 2)
{
printf("No solution!
");
continue;
}
int cha = c - 'a';
int st = (mid - r - 1) / 2;//st*2+2=mid-r+1
int ed = (mid + r - 3) / 2;//ed*2+2=mid+r-1
printf("%d %d
", st, ed);
for (int i = mid - r + 1; i <= mid + r - 1; i += 2)
{
s[i] -= cha;
if (s[i] < 'a')
s[i] += 26;
printf("%c", s[i]);
}
printf("
");
}
return 0;
}