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HDU3294:Girls' research

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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:给出一个字母,例如b,那么b就是真正的a,c就是真正的b,再给出一个
字符串,让你求出最大回文串
思路:先打出字母表,然后分奇偶性找出最大的回文串
*/
#include <stdio.h>
#include <string.h>

char str[200005];
char word[30];//记录该字母真正代表的字母
int main()
{
    int i,j,len,head,end,max;
    char c,ch;
    while(~scanf("%c%s",&c,str))
    {
        getchar();
        max=0;
        len=strlen(str);
        j = c-'a';//记录相差
        ch = 'a';
        head = end = 0;
        for(i = j; i<j+26; i++)
            word[i%26] = ch++;//字母表
        for(i=0; i<len; i++)
        {
            for(j=0; i-j>=0&&i+j<len; j++)//奇数长度的回文串
            {
                if(str[i-j]!=str[i+j])
                    break;
                if((2*j+1)>max)
                {
                    max=2*j+1;//最大长度
                    head=i-j;//开始位置
                    end=i+j;//末位置
                }
            }
            for(j=0; i-j>=0&&i+j+1<len; j++)//偶数长度的回文串
            {
                if(str[i-j]!=str[i+j+1])
                    break;
                if((2*j+2)>max)
                {
                    max=2*j+2;
                    head=i-j;
                    end=i+j+1;
                }
            }
        }
        if(end - head<2)
            printf("No solution!\n");
        else
        {
            printf("%d %d\n",head,end);
            for(i=head; i<=end; i++)
                printf("%c",word[str[i]-'a']);
            printf("\n");
        }
    }
    return 0;
}

作者:libin56842 发表于2013-5-5 20:42:44 原文链接
阅读:34 评论:0 查看评论

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