CodeForces - 1606A AB Balance

Posted 2022-06-23 海岛Blog

A. AB Balance
time limit per test 2 seconds
memory limit per test 256 megabytes

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
input
4
b
aabbbabaa
abbb
abbaab
output
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.

问题链接：CodeForces - 1606A AB Balance
问题简述：（略）
问题分析：（略）

AC的C语言程序如下：

/* CodeForces - 1606A AB Balance */

#include <stdio.h>
#include <string.h>

#define N 100 + 1
char s[N];

int main()

    int t;
    scanf("%d", &t);
    while (t--) 
        scanf("%s", s);

        s[0] = s[strlen(s) - 1];

        puts(s);
    

    return 0;