Codeforces 464 A. No to Palindromes!

因为初始的串是不包含回文的,所以只要判断最后一个字符和前一个和前前一个不要构成回文就可以了,从最后一位开始加1枚举,如果不可以就往前枚举一位,一直到可以把当前的字符确定下来为止,再往后面构造尽量小的不构成回文的串就可以了.

A. No to Palindromes!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and sdoesn't contain any palindrome contiguous substring of length 2 or more.

Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.

Input

The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting of nsmall English letters. It is guaranteed that the string is tolerable (according to the above definition).

Output

If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).

Sample test(s)

input

3 3cba

output

NO

input

3 4cba

output

cbd

input

4 4abcd

output

abda

Note

String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.

The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.

A palindrome is a string that reads the same forward or reversed.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,p;char str[1111];int main(){    cin>>n>>p;    cin>>str;    int mark=-1;    bool flag=false;    char head=str[0];    for(int i=n-1;i>=0;i--)    {        if(flag==true) break;        for(int j=str[i]-'a'+1;j<p;j++)///枚举最后面的        {            if(flag==true) break;            if(i-1>=0&&i-2>=0)            {                if(j!=str[i-1]-'a'&&str[i-2]!=j+'a')                {                    mark=i; str[mark]=j+'a';                    flag=true; break;                }            }            else if(i-1>=0)            {                if(j!=str[i-1]-'a')                {                    mark=i; str[mark]=j+'a';                    flag=true; break;                }            }        }        if(flag==false)            str[i]='*';    }    if(mark==-1)    {        if(head-'a'+1>=p)        {            cout<<"NO"<<endl;            return 0;        }        else        {             mark=0; str[0]=head+1;        }    }    flag=true;    for(int i=mark+1;i<n;i++)    {        bool temp=false;        for(int j=0;j<p;j++)        {            char ch=j+'a';            if(i-1>=0&&i-2>=0)            {                if(ch!=str[i-1]&&ch!=str[i-2])                {                    str[i]=ch;                    temp=true;                }            }            else if(i-1>=0)            {                if(ch!=str[i-1])                {                    str[i]=ch;                    temp=true;                }            }            if(temp==true) break;        }        if(temp==false)        {            flag=false;            break;        }    }    if(flag)    {        cout<<str<<endl;    }    else    {        cout<<"NO"<<endl;    }    return 0;}