题目描述

A string $s$ of length $n$ ( $1 \le n \le 26$ ) is called alphabetical if it can be obtained using the following algorithm:

  • first, write an empty string to $s$ (i.e. perform the assignment $s$ := "");
  • then perform the next step $n$ times;
  • at the $i$ -th step take $i$ -th lowercase letter of the Latin alphabet and write it either to the left of the string $s$ or to the right of the string $s$ (i.e. perform the assignment $s$ := $c+s$ or $s$ := $s+c$ , where $c$ is the $i$ -th letter of the Latin alphabet).

In other words, iterate over the $n$ first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string $s$ or append a letter to the right of the string $s$ . Strings that can be obtained in that way are alphabetical.

For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".

From the given string, determine if it is alphabetical.

输入格式

The first line contains one integer $t$ ( $1 \le t \le 10^4$ ) — the number of test cases. Then $t$ test cases follow.

Each test case is written on a separate line that contains one string $s$ . String $s$ consists of lowercase letters of the Latin alphabet and has a length between $1$ and $26$ , inclusive.

输出格式

Output $t$ lines, each of them must contain the answer to the corresponding test case. Output YES if the given string $s$ is alphabetical and NO otherwise.

You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).

输入输出样例

输入 #1

11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba

输出 #1

YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO

说明/提示

The example contains test cases from the main part of the condition.

分析

要判断字符串是否是通过按字典序依次在前后累加字母得到的,我们只需要按倒字典序依次扫描字符串的第一个和最后一个字母,如果第一个和最后一个字母都不是当前按倒字典序轮询到的字母,则该字符串不是通过按字典序依次在前后累加字母得到的。

因为 $n \le 26$,所以任意字母都不可能出现两次,因此我们只需要将 $26$ 个字母轮询一遍,代码实现较为简单。

代码

#include<bits/stdc++.h>
using namespace std;
int t,l,r,i,flag;
char a[30];
char ch[27]={0,'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
int main(){
    scanf("%d",&t);
    while(t--){
        cin>>a+1;
        l=1,flag=0;
        i=r=strlen(a+1);
        while(l<=r){
            if(a[l]==ch[i]){
                l++;i--;
                continue;
            }
            if(a[r]==ch[i]){
                r--;i--;
                continue;
            }
            flag=1;
            printf("NO\n");
            break;
        }
        if(!flag)printf("YES\n");
    }
    return 0;
}
最后修改:2021 年 07 月 20 日 10 : 43 AM
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