A. ABC String.

原题链接：[https://codeforces.com/contest/1494/problem/A]

题面

You are given a string a, consisting of n characters, n is even. For each i from 1 to n ai is one of ‘A’, ‘B’ or ‘C’.

A bracket sequence is a string containing only characters “(” and “)”. A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters “1” and “+” between the original characters of the sequence. For example, bracket sequences “()()” and “(())” are regular (the resulting expressions are: “(1)+(1)” and “((1+1)+1)”), and “)(”, “(” and “)” are not.

You want to find a string b that consists of n characters such that:

b is a regular bracket sequence;
if for some i and j (1≤i,j≤n) ai=aj, then bi=bj.
In other words, you want to replace all occurrences of ‘A’ with the same type of bracket, then all occurrences of ‘B’ with the same type of bracket and all occurrences of ‘C’ with the same type of bracket.

Your task is to determine if such a string b exists.

Input
The first line contains a single integer t (1≤t≤1000) — the number of testcases.

Then the descriptions of t testcases follow.

The only line of each testcase contains a string a. a consists only of uppercase letters ‘A’, ‘B’ and ‘C’. Let n be the length of a. It is guaranteed that n is even and 2≤n≤50.

Output
For each testcase print “YES” if there exists such a string b that:

b is a regular bracket sequence;
if for some i and j (1≤i,j≤n) ai=aj, then bi=bj.
Otherwise, print “NO”.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).

Example
input

4
AABBAC
CACA
BBBBAC
ABCA

output
YES
YES
NO
NO

Note
In the first testcase one of the possible strings b is “(())()”.

In the second testcase one of the possible strings b is “()()”.

题解

题意中明确指出找出正则的括号序列。例如，括号序列“（）（）”和“（（））”是正则的（结果表达式是：”（1）+（1）”和“（（1+1）+1）”），而“）（”、“（”和“）”不是。

所以第一种情况每次输入的字符串中第一个字母和最后一个字母不能相同，相同的话则不能组成括号序列例如ABA会形成‘（）（’ 或者）（）。

如果头字母与尾字母不同时，查找以头字母为（另外不同字母为）数量是否相同，若不同再查找以尾字母为）另外不同字母为（数量是否相同。

代码

#include<iostream>
#include<string.h>
using namespace std;int main()
{int n;cin>>n;while(n--){string a;cin>>a;char s1,s2;int num1=0,num2=0;int f=1;s1=a[0];s2=a[a.size()-1];if(s1==s2)f=0;else{for(int i=0;i<a.size();i++){if(a[i]==s1)num1++;elsenum2++;if(num1<num2){f=0;break;}}if(num1!=num2)f=0;if(f==0){int num3=0,num4=0;for(int i=0;i<a.size();i++){if(a[i]==s2)num3++;elsenum4++;if(num4<num3){f=0;break;}}}}if(f==1)cout<<"YES"<<endl;elsecout<<"NO"<<endl;}return 0;
}```

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A. ABC String.

A. ABC String.

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