CodeForces - 665C Simple Strings （水题）

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这道题目的目标是将输入的字符串转换为简单字符串，即相邻字符均不相同。题目提供了一种策略，对于每三个连续字符，如果中间的字符与两边相同，则替换为不同的字母。需要找到使字符串变为简单字符串所需的最小修改次数。

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题目链接：http://codeforces.com/problemset/problem/665/C点击打开链接

zscoder loves simple strings! A string t is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoderare 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

对每三个数的中间那个数判断看左右跟左右是否相等相等就改成跟那两个数不一样的

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits.h>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#define maxn 101000
using namespace std;
int main()
{
    string s;
    cin >> s;
    for(int i=1;i<s.length();i++)
    {
        if(s[i-1]==s[i])
        {
            for(int ii=0;ii<=3;ii++)
                if(s[i-1]!=ii+'a'&&s[i+1]!=ii+'a')
                {
                    s[i]=ii+'a';
                    break;
                }
        }
    }
    cout << s;
}