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  • 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 ababa,zscoder are simple whereas aaadd 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;
    }


    
    

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  • 原文地址:https://www.cnblogs.com/dacc123/p/8228676.html
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