hdu3294---Girls' research

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 babda abcd

 

Sample Output

0 2azaNo solution!
题目大意：输入多组一个字符和一串字符，仅由小写字母组成
例如 c abcba，c代表把串中的c改成a，d改成b... b改成z，a改成y...
求最长回文子串，输出起始和终止下标，并输出回文子串，回文串长度是1输出No
代码：
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define N 200009char s[N],t[N<<1];int p[N<<1];int main(){    char c;    int j,i;    while(~scanf(" %c",&c))    {        scanf("%s",s);        int n=c-'a';        int len=strlen(s);        for(i=0; i<len; i++)        {            s[i]=s[i]-n;            if(s[i]<'a')                s[i]=s[i]+26;        }        for(i=1; i<=len; i++)            t[i*2]=s[i-1],t[2*i+1]='#';        t[0]='!',t[1]='#';        len=2*len+1;        t[len+1]='\0';        int id=0,mx=0,ans=0,l,r;        for(i=1; i<=len; i++)        {            if(mx>i) p[i]=min(p[2*id-i],mx-i);            else p[i]=1;            while(t[i-p[i]]==t[i+p[i]]) p[i]++;            if(i+p[i]>mx) mx=i+p[i],id=i;            if(ans<p[i])            {                l=(i-p[i])/2;//起始位置                r=l+p[i]-2;//终止位置                ans=p[i];//如果写ans=p[i]-1则此if判断条件和下方if判断条件都需改            }        }        if(ans==2)            puts("No solution!");        else        {            printf("%d %d\n",l,r);            for(j=l; j<=r; j++)                cout<<s[j];            cout<<endl;        }    }    return 0;}