CodeForces - 665C Simple Strings（贪心）

最新推荐文章于 2017-09-07 13:49:00 发布

原创最新推荐文章于 2017-09-07 13:49:00 发布 · 408 阅读

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本文介绍了一个简单的算法，用于将任意给定的字符串转换为每个相邻字符都不相同的简单字符串。通过扫描字符串并替换重复字符，确保结果字符串满足条件，且替换次数最少。

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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

长教训了，不看题就写，wa的不要不要的

受样例误导，以为代替的字母必须出现过，然而并不是，那么我只要有abc就可以妥妥的解决问题了好不好

扫一遍，当前字母若与其前一个字母相同，那么没商量肯定得换掉，换的那个字母只要和前一个字母和后一个字母都不同就好了

代码

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<queue>
using namespace std;
char s[200005];
int main()
{
    while(~scanf("%s",s))
    {
        int n=strlen(s);
        for(int i=1;i<n;i++)
        {
            if(s[i]==s[i-1])
            {
                int temp=0;
                if(s[i]-'a'==temp)temp++;
                if(s[i+1]-'a'!=temp)s[i]=temp+'a';
                else
                {
                    temp++;
                    if(s[i-1]-'a'==temp)temp++;
                    s[i]=temp+'a';
                }
            }
        }
        printf("%s\n",s);
    }
    return 0;
}