Girls' research---hdu3294（回文子串manacher）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N = 4e5+7;

int p[N], id, mx, ans, Index;
char s[N], ch[5], s0[N];

void Manacher(char s[], int n)
{
    id = mx = ans = 0;
    for(int i=2; i<n; i++)
    {
        if(mx>i)
            p[i] = min(p[id*2-i], mx-i);
        else
            p[i] = 1;
        while(s[i-p[i]] == s[i+p[i]])
            p[i]++;
        if(p[i]+i>mx)
        {
            mx = p[i]+i;
            id = i;
        }
        if(p[i]>ans)
        {
            ans = p[i];
            Index = i;///记录结果的中心下标；
        }
    }
}

int main()
{
    while(scanf("%s%s", ch, s)!=EOF)
    {
        int len = strlen(s);
        int k = ch[0]-‘a‘;
        for(int i=0; i<len; i++)///转化成真正的串s；
        {
            if(s[i]-k<‘a‘)
                s[i]=s[i]-k+26;
            else
                s[i]=s[i]-k;
        }
        ///printf("%s\n", s);
        for(int i=len; i>=0; i--)
        {
            s[i+i+2] = s[i];
            s[i+i+1] = ‘#‘;
        }
        s[0] = ‘$‘;
        Manacher(s, 2*len+2);
        ans = ans-1;
        k = ans/2;///前面或后面有几个字母在回文串中；
        if(k == 0)
        {
            printf("No solution!\n");
            continue;
        }
        int L = Index-ans+1;
        int R = Index+ans-1;
        printf("%d %d\n", (L-2)/2, (R-2)/2);
        for(int i=L; i<=R; i++)
            if(s[i]!=‘#‘)
                printf("%c", s[i]);
        printf("\n");
    }
    return 0;
}