官方题解
https://codeforces.com/blog/entry/89319

A

题意:给定一个串，能否通过若干次添加’a’操作使之变成非回文串，能顺带输出结果
解法:当且仅当给定串全为’a’时必为回文串，其他情况则必可通过加’a’操作构造成非回文串
构造方式：分别向头/尾加’a’,两者比存在非回文答案

AC代码

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; ++i)
#define per(i, l, r) for (int i = l; i >= r; --i)
#define mset(s, _) memset(s, _, sizeof(s))
#define mcpy(s, _) memcpy(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define vi vector<int>
#define vpii vector<pii> 
#define mp(a, b) make_pair(a, b)
#define debug system("pause")
#define pll pair <ll, ll>
#define fir first
#define sec second
#define inf 0x3f3f3f3f
#define pline cout << endl << endl 
inline int lowbit(int x) {return x & -x;} 

const int N = 1e6 + 10, mod = 1e9 + 7;
int t, n;

int main() {
    cin >> t;
    while(t -- ) {
        string str; cin >> str;
        int num = str.size();
        int numa = 0; 
        rep(i, 0, num - 1) {
            if(str[i] == 'a') numa ++ ;
        }
        if(numa == num) {
            puts("NO"); continue;
        }
        string t1, t2 = "a";
        t1 = str + "a"; string t11(t1); reverse(t11.begin(), t11.end());
        t2 = t2 + str; string t22(t2); reverse(t22.begin(), t22.end()); 
        if(t11 != t1) {
            cout << "YES" << endl; cout << t1 << endl;
            continue;
        } else {
            puts("YES"); cout << t2 << endl;
        }

    }

    return 0; 
}   

B

题意:给定01串，每次能自由将0， 1个数相同的前缀进行取反操作，问能够通过有限次操作将当前串变成目标串
解法:通过观察易得对于0,1个数相同且与目标串匹配的前缀，取反操作后定能复原，而对于0,1串个数不同但是与目标串匹配的前缀，取反操作以后定不能复原,对于不能与目标串匹配的前缀，如果0,1个数相等，则必能通过若干操作将这部分前缀变成目标串前缀，故只需动态的分别统计与目标串匹配和不匹配的前缀的0, 1个数是否相等判断结果即可
AC代码

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; ++i)
#define per(i, l, r) for (int i = l; i >= r; --i)
#define mset(s, _) memset(s, _, sizeof(s))
#define mcpy(s, _) memcpy(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define vi vector<int>
#define vpii vector<pii> 
#define mp(a, b) make_pair(a, b)
#define debug system("pause")
#define pll pair <ll, ll>
#define fir first
#define sec second
#define inf 0x3f3f3f3f
#define pline cout << endl << endl 
inline int lowbit(int x) {return x & -x;} 

const int N = 1e6 + 10, mod = 1e9 + 7;
int t, n;

int main() { 
    cin >> t;
    while(t -- ) {
        cin >> n; string a, b; cin >> a >> b;
        if(a == b) {
            puts("YES"); continue;
        }
        int p1 = 0, p0 = 0, up1 = 0, up0 = 0;
        bool fg = 1, flag = 0;
        for(int i = 0; i < n; ) {
            if(a[i] == b[i]) {
                flag = 1;
                if(i - 1 >= 0 && a[i - 1] != b[i - 1]) {
                    if(p1 != p0 || up1 != up0) {
                        fg = 0; break;
                    }
                    p0 += up0; p1 += up1; up0 = 0; up1 = 0; 
                }

                if(a[i] == '0') p0 ++ ;
                else p1 ++ ;
                i ++ ;
            }
            else {
                flag = 0;
                if(a[i] == '0') up0 ++ ;
                else up1 ++ ;
                i ++ ;
            }
        }
        if(flag == 0 && (p1 != p0 || up1 != up0)) fg = 0;
        if(fg) puts("YES");
        else puts("NO");
    } 

    return 0; 
}   

C(构造)

题意:给定01串s，能否构造出一组合法括号串A, B使得当s[i] == ‘1’时，A[i] == B[i], 当s[i] == ‘0’时，A[i] != B[i]
解法:1.要保证序列的合法，首先要保证第一个字符必定时’(‘，最后一个字符必定时’)’，故s[0] 与 s[n - 1] 必须为‘1’
2.当01串中‘0’的个数为奇数时，必不存在解

因为给定的n为偶数，对于两个串总体而言，'(' 与 ')'的个数分别为n
对于01串中的'1'，若‘1’的个数为奇数，因为A，B中若s[i] == 1，则A[i] == B[i], 故'1'对应下标的'('和')'的个数都必为偶数, '1'的个数为偶数更显然

对于01串中的'0'，若’0‘的个数为奇数，因为A，B中若s[i] == 0，则A[i] != B[i], 所以’0‘对应下标的'('与')'的个数必为奇数，这将使得两个串的'(' 与 ')'的个数为奇数，不合题意，故排除

对于0的个数非奇数非奇数的情况，我们可以如下构造出答案，统计出1的个数k，将1的前k/2的位置设置为(, 后k/2位置设置为), 对于0下标对应的位置则一次设置为( , )
[注]:此答案可通过分离1， 0的方式证明必定使得A, B都合法

AC代码

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; ++i)
#define per(i, l, r) for (int i = l; i >= r; --i)
#define mset(s, _) memset(s, _, sizeof(s))
#define mcpy(s, _) memcpy(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define vi vector<int>
#define vpii vector<pii> 
#define mp(a, b) make_pair(a, b)
#define debug system("pause")
#define pll pair <ll, ll>
#define fir first
#define sec second
#define inf 0x3f3f3f3f
#define pline cout << endl << endl 
inline int lowbit(int x) {return x & -x;} 

const int N = 1e6 + 10, mod = 1e9 + 7;
int t, n;

int main() { 
    cin >> t;
    for(int k = 1; k <= t; k ++ ) {

        cin >> n;
        string str; cin >> str;
        if(t == 64) cout << str << endl;
        if(str[0] != '1' || str[n - 1] != '1') {
            puts("NO"); continue;
        }
        int num0 = 0, num1 = 0;
        rep(i, 0, n - 1) {
            if(str[i] == '0') num0 ++ ;
            else num1 ++ ; 
        }
        if(num0 % 2) {
            puts("NO"); continue;
        }

        puts("YES");
        string a, b;
        int s1 = 0, s0 = 0, t = 0;
        rep(i, 0, n - 1) {
            if(str[i] == '1') s1 ++ ;
            else s0 ++ ;
            if(str[i] == '1' && s1 <= num1 / 2) {a += '('; }
            else if(str[i] == '1' && s1 > num1 / 2) {a += ')'; }
            else if(str[i] == '0' && t == 0) {a += '('; t = 1;}
            else {a += ')'; t = 0;}
        }
        rep(i, 0, n - 1) {
            if(str[i] == '1') b += a[i];
            else {
                if(a[i] == '(') b += ')';
                else b += '(';
            }
        }
        cout << a << endl << b << endl;
    }

    return 0; 
}