【题解】Alphabetical Strings

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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