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Girls‘ research

b babd
a abcd

0 2
aza
No solution!

题意:给出一个串和一个字符ch,这个串是原串把a替换成ch,b换成ch+1,c换成

ch+2..得来的,然后求原串的最长回文子序列.

manachar以后遍历一遍数组就好了,注意+26以后取模.

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
#define maxn 211111

int len[maxn<<1];
char s[maxn<<1];
char a[maxn];
char change;

int f (int x) {
    return (x-1)>>1;
}

int manachar (char *p) {
    int n = strlen (p), l = 0;
    s[l++] = '@';
    s[l++] = '#';
    for (int i = 0; i < n; i++) {
        s[l++] = p[i];
        s[l++] = '#';
    }
    s[l] = 0;

    int Max = 0, pos = 0, ans = 0;
    for (int i = 1; i < l; i++) {
        if (Max > i) {
            len[i] = min (len[2*pos-i], Max-i);
        }
        else
            len[i] = 1;
        while (s[i+len[i]] == s[i-len[i]])
            len[i]++;
        ans = max (ans, len[i]);
        if (len[i]+i > Max) {
            Max = len[i]+i;
            pos = i;
        }
    }
    ans--;
    if (ans == 1) {
        printf ("No solution!\n");
        return 0;
    }
    for (int i = 1; i < l; i++) {
        if (len[i]-1 == ans) {
            printf ("%d %d\n", f(i-len[i]+2), f(i+len[i]-2));
            for (int j = i-len[i]+1; j <= i+len[i]-1; j++) if (s[j] != '#') {
                printf ("%c", 'a'+(s[j]-change+26)%26);
            }
            puts ("");
            return 0;
        }
    }
    return 0;
}

int main () {
    //freopen ("in.txt", "r", stdin);
    while (scanf ("%s%s", s, a) == 2) {
        change = s[0];
        manachar (a);
    }
    return 0;
}

HDU 3294 (manachar)

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原文地址：http://blog.csdn.net/morejarphone/article/details/51366170