【CODEFORCES】 C. No to Palindromes!

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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 ofn 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 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, ..., s_i = t_i, s_i + 1 > t_i + 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.

题解：这道题看似简单，实际上出的还是很好的。

一开始用暴力，发现在第20多个点超时之后才想到优化一下。

一开始从最低位开始加，然后返回此次过程中改变的最高位的位置i，然后从第i位向后检查第i位是否和i-1位，i-2位相等，如果相等，第i位就继续a[i]++，然后再判断………………

这样如此往复，我们可以让add(i)表示执行a[i]++（并且同时进位）并返回所改变的最高位的值，check(i)表示从第i位向后检查是否合法，如果不合法就返回不合法的位置，如果合法就输出解。

注意：add(i)和check(i)是一个循环往复的过程。

#include <iostream>#include <cstring>#include <cstdio>using namespace std;char s[1002],a[1002];int n,p,i,x;int add(int k){    int i=k,t;    a[i]++;  t=0;    while (a[i]-96>p && i>=0)    {        t=1;        a[i]='a';        i--;        a[i]+=t;    }    return i;}int _check(int k){    for (int i=k;i<=n-1;i++)        if (a[i]==a[i-1] || a[i]==a[i-2]) return i;    return n;}int main(){    scanf("%d%d",&n,&p);    scanf("%s",&s);    for (int i=0;i<=n;i++) a[i]=s[i];    i=n-1;    while (1)    {        x=add(i);        if (x==-1)        {            cout <<"NO"<<endl;            return 0;        }        i=_check(x);        if (i==n)        {            printf("%s",a);            return 0;        }    }    return 0;}

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