CodeForeces 665C Simple Strings

C. Simple Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

zscoder loves simple strings! A string t is
called simple if every pair of adjacent characters are distinct. For example ab, aba,zscoder are
simple whereas aa, add are not simple.

zscoder is given a string s. He
wants to change a minimum number of characters so that the string s becomes simple. Help him with this task!

Input

The only line contains the string s (1 ≤ |s| ≤ 2·105)
— the string given to zscoder. The string s consists
of only lowercase English letters.

Output

Print the simple string s' — the string s after
the minimal number of changes. If there are multiple solutions, you may output any of them.

Note that the string s' should also consist of only lowercase English letters.

Examples

input

aab

output

bab

input

caaab

output

cabab

input

zscoder

output

zscoder

把重复的区间找出来，隔一个变一个。

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 100000
int len[2*MAX+5];
char a[2*MAX+5];
int main()
{
    scanf("%s",a);
    int l=strlen(a);
    len[l-1]=1;
    for(int i=l-2;i>=0;i--)
    {
        len[i]=1;
        if(a[i]==a[i+1])
            len[i]+=len[i+1];
    }
    for(int i=0;i<l;i)
    {
        if(len[i]!=1)
        {
            for(int j=i+1;j<=i+len[i]-1;j+=2)
            {
                if(j==i+len[i]-1)
                {
                    for(int p=0;p<26;p++)
                    {
                        if(('a'+p)==a[i]||('a'+p)==a[j+1])
                            continue;
                        a[j]='a'+p;
                        break;
                    }
                }
                else
                   a[j]=(a[i]=='z'?a[i]-1:a[i]+1);
            }
            i=i+len[i];
        }
        else
            i++;
    }
    for(int i=0;i<l;i++)
    {
        printf("%c",a[i]);
    }

    cout<<endl;
    return 0;
}