Codeforces Round #265 (Div. 2) C 暴力+ 找规律+ 贪心

C. 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 s 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, ..., 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.

题解转自：http://www.cnblogs.com/iwtwiioi/p/3961709.html

题意：给你一个没有长度为2以上（包括2）的回文串的一个字符串，然后让你求比他大的所有排列且没有长度为2以上（包括2）的下一个最小的排列。

剩下的1个半小时都在搞这题。。。。。。。。。。T_T

神题。。

我dfs写渣

判重写渣

然后滚粗。

这里有个贪心。

串的任意部分都不能是回文=串中任意连续3个字符互不相同

那么如果从左往右找，那么只需要判断x-1和x-2与x不等即可

其次我们要保证字典序最小，所以我们从右向左依次尝试修改，只要找到不符合回文条件的点，就修改它。这样保证了前边的最小。。

而且能证明，后边一定能构造出一个不是回文的串。例如m=4，我可以构造abcabcabc。。。。。。。。。。。。（x-1和x-2与x不等）

程序中有访问 char 数组的负下标，感觉太高大上了,我试了下，负下标的%d输出为0

网上有人说原因是  ：记住，下标只是个操作符。 x[n] 等价于 *(x + n)

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>

#define N 1005
#define M 15
#define mod 6
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int n,p;
char s[N];

char ok(int t)
{
    char i;
    for(i=s[t]+1;i-'a'+1<=p;i++){
        if(s[t-1]!=i && s[t-2]!=i) return i;
    }
    return 0;
}

int solve()
{
    int i;
    char t;
   // if(p==1) return 0;
   // if(n==1) return
    for(i=n-1;i>=0;i--){
        t=ok(i);
        if( t!=0 ){
            s[i]=t;
            break;
        }
    }
    if(i==-1) return 0;

    i++;
    for(;i<n;i++){
        s[i]='a'-1;
        t=ok(i);
        if( t!=0 ){
            s[i]=t;
        }
    }
    return 1;
}

int main()
{
    //freopen("data.in","r",stdin);
    //scanf("%d",&T);
    //for(int cnt=1;cnt<=T;cnt++)
    //while(T--)
    while(scanf("%d%d",&n,&p)!=EOF)
    {

        scanf("%s",s);
       // if(p==1){
        //    printf("NO
");continue;
       // }
        if(solve()==0){
            printf("NO
");
        }
        else{
            printf("%s
",s);
        }
    }

    return 0;
}