CF 464 A No to Palindromes! 最後の後ろの可変アルファベットを見つけます.

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 ands doesn'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 n small 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 3
cba

output

NO

input

3 4
cba

output

cbd

input

4 4
abcd

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.

/*  
         ；    ，         ，                   ，       m；
            ，              ；

       ；
          。             a      ，
      ，           ；
*/  
#include<stdio.h>
int n,m;
char a[2000];
bool judge(char ch,int i)
{
	for(;ch<'a'+m;ch++)
	{
		if((i==0||ch!=a[i-1])&&(i<=1||ch!=a[i-2]))
		{
			a[i]=ch;
			return 1;
		}
		
	}
	return 0;
}
void get(int i)
{
	for(i++;i<n;i++)
	{
		for(a[i]='a';a[i]<'a'+m;a[i]++)
		{
			if((a[i]!=a[i-1]||i==0)&&(i<=1||a[i]!=a[i-2]))
				break;
		}
	}
}
	
int main()
{
	int ok,i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		scanf("%s",a);
		ok=-1;
		for(i=n-1;i>=0;i--)
		{
			if(judge(a[i]+1,i))
			{
				ok=i;
				break;
			}
		}
		if(ok!=-1)
		{
			get(ok);
			puts(a);
		}
		else
		{
			puts("NO");
		}
	}
	return 0;
}