A. AB Balance

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a string s of length n consisting of characters a and/or b.

Let AB(s) be the number of occurrences of string ab in s as a substring. Analogically, BA(s) is the number of occurrences of ba in s as a substring.

In one step, you can choose any index i and replace si with character a or b.

What is the minimum number of steps you need to make to achieve AB(s)=BA(s)?

Reminder:

The number of occurrences of string d in s as substring is the number of indices i (1≤i≤|s|−|d|+1) such that substring sisi+1…si+|d|−1 is equal to d. For example, AB(aabbbabaa)=2 since there are two indices i: i=2 where aabbbabaa and i=6 where aabbbabaa.

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤1000). Description of the test cases follows.

The first and only line of each test case contains a single string s (1≤|s|≤100, where |s| is the length of the string s), consisting only of characters a and/or b.

Output
For each test case, print the resulting string s with AB(s)=BA(s) you’ll get making the minimum number of steps.

If there are multiple answers, print any of them.

Example
inputCopy
4
b
aabbbabaa
abbb
abbaab
outputCopy
b
aabbbabaa
bbbb
abbaaa
Note
In the first test case, both AB(s)=0 and BA(s)=0 (there are no occurrences of ab (ba) in b), so can leave s untouched.

In the second test case, AB(s)=2 and BA(s)=2, so you can leave s untouched.

In the third test case, AB(s)=1 and BA(s)=0. For example, we can change s1 to b and make both values zero.

In the fourth test case, AB(s)=2 and BA(s)=1. For example, we can change s6 to a and make both values equal to 1.

题目大意：有一串由a，b组成的字符串，一次操作可以将a换成b或者将b换成a，至少多少次操作可以是字符串中的ab字串的数量等于ba字串的数量，输出的是变换后的字符串

思路：因为是有a，b组成的所以，ab的数量与ba的数量要不相同，要不相差一，如果字符串以a开头那么，这个字符串中的n(ab)>=n(ba);所以如果n(ab)==n(ba)，直接输出原始串，如果n(ab)>n(ba),把开头的a换成b就可以，然后在输出，b开头同理

代码：

#include <iostream>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        string s;
        cin>>s;
        int a=0,b=0;
        for(int i=1;i<s.size();i++)
        {
            if(s[i]=='a'&&s[i-1]=='b') b++;
            else if(s[i]=='b'&&s[i-1]=='a') a++;
        }
        if(a==b) cout<<s<<endl;
        else if(a>b)
        {
            if(s[0]=='a') s[0]='b';
            cout<<s<<endl;
        }
        else if(b>a)
        {
            if(s[0]=='b') s[0]='a';
            cout<<s<<endl;
        }
    }
    return 0;
}