Codeforces 464A. No to Palindromes! (dp,向前推理,向后查询)

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 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, ..., 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.

http://codeforces.com/contest/464/problem/A

//Hello. I'm Peter.#include<cstdio>#include<iostream>#include<sstream>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<algorithm>#include<functional>#include<cctype>#include<ctime>#include<stack>#include<queue>#include<vector>#include<set>#include<map>using namespace std;typedef long long ll;typedef long double ld;#define peter cout<<"i am peter"<<endl#define input freopen("data.txt","r",stdin)#define INT (0x3f3f3f3f)*2#define LL (0x3f3f3f3f3f3f3f3f)*2#define gsize(a) (int)a.size()#define len(a) (int)strlen(a)#define slen(s) (int)s.length()#define pb(a) push_back(a)#define clr(a) memset(a,0,sizeof(a))#define clr_minus1(a) memset(a,-1,sizeof(a))#define clr_INT(a) memset(a,INT,sizeof(a))#define clr_true(a) memset(a,true,sizeof(a))#define clr_false(a) memset(a,false,sizeof(a))#define clr_queue(q) while(!q.empty()) q.pop()#define clr_stack(s) while(!s.empty()) s.pop()#define rep(i, a, b) for (int i = a; i < b; i++)#define dep(i, a, b) for (int i = a; i > b; i--)#define repin(i, a, b) for (int i = a; i <= b; i++)#define depin(i, a, b) for (int i = a; i >= b; i--)#define pi 3.1415926535898#define eps 1e-6#define MOD 1000000007#define MAXN 100100#define M 200int n,p;char c,to;string row,ans;bool is_true_ok(string s,int pos){    bool ok,is_ok=true;    int len=slen(s);    rep(i,pos+1,len)    {        ok=false;        for(c='a';c<=to;c++)        {            if(i==0 || (i>0&&c!=s[i-1]))            {                if(i==1 || (i>1&&c!=s[i-2]))                {                    ok=true;                    s[i]=c;                    break;                }            }        }        if(!ok)        {            is_ok=false;            break;        }    }    if(!is_ok) return false;    else    {        ans=s;        return true;    }}int main(){//    input;    string s;    bool ok;    cin>>n>>p;    to=(p-1)+'a';    cin>>row;    depin(i,n-1,0)    {        ok=false;        for(c=row[i]+1;c<=to;c++)        {            if(i==0 || (i>0&&c!=row[i-1]))            {                if(i==0 || i==1 || (i>1&&c!=row[i-2]))                {                    ok=true;                    break;                }            }        }        if(ok)        {            s=row;            s[i]=c;            if(is_true_ok(s,i))            {                cout<<ans<<endl;                exit(0);            }        }    }    cout<<"NO"<<endl;}