$$$x\oplus x=0$$$

$$$x\oplus x\oplus x=x$$$

The answer depends on parity.

If $$$m$$$ is even, then the result of the formula is always $$$0$$$. Otherwise, it can be any positive integer, because all the elements in the sequence $$$a$$$ is positive.

#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int n, m;
void run()
{
    scanf("%d%d", &n, &m);
    puts((m & 1) ^ (!n) ? "Yes" : "No");
}
int main()
{
    int T = 1;
    scanf("%d", &T);
    while (T--)
        run();
}

Lexicographically smallest.

If you can keep the current character a, then do it.

Since a is the lexicographically smallest character, then it's always optimal to insert a instead of others.

Use # in the string $$$b$$$ greedily.

#include <iostream>
#include <string>
using namespace std;
const int N = 1e5 + 10;
int n, m, cts, idx;
string s, t;
int main()
{
    cin >> n >> m >> s >> t;
    for (int i = 0; i < n; i++)
    {
        cts += (t[i] == '#');
    }
    for (auto &i : s)
    {
        if (i != '#')
            continue;
        i = idx + 'a';
        if (cts >= 25 - idx)
        {
            cts -= 25 - idx;
            idx = 0;
            continue;
        }
        idx++;
        idx = (idx == 26 ? 0 : idx);
    }
    cout << s << '\n';
}

Denote $$$c_{i,x}$$$ as the total number moments when exactly $$$x$$$ roses are blooming when or before the moment $$$i$$$.

If we have a virtual rose which blooms in the range $$$[1,i]$$$, then how to denote the number the answer will increase when we move that flower's blooming time to an infinitely faraway place?

The answer of the question raised in hint $$$1$$$ is $$$\sum_{j=2}^\infty c_{i,j}+c_{i,2}$$$.

Prefix sum.

We can calculate the sequence $$$c$$$ at each turning point. We call a moment a turning point when a rose blooms or fades at this moment.

Then, calculate value in hint $$$2$$$ for each turning point.

It can be proved that moving $$$[x,y]$$$ away is equivalent to move $$$[1,y]$$$ away then move $$$[1,x-1]$$$ back.

So use prefix sum to calculate answer for each rose.

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2e5 + 10;
using ll = long long;
int n, m, x, idx, cur;
ll mx, res[4], p2[N];
using pii = pair<int, int>;
pii dst[N << 2];
bool mp[N];
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &x);
        dst[++idx] = {x, i};
        dst[++idx] = {x + m, i};
    }
    sort(dst + 1, dst + idx + 1);
    for (int i = 1; i <= idx; i++)
    {
        auto &[a, b] = dst[i];
        res[min(cur, 3)] += a - dst[i - 1].first;
        if (!mp[b])
        {
            cur++;
            mp[b] = true;
            p2[b] = res[2] * 2 + res[3];
            continue;
        }
        mx = max(mx, res[2] * 2 + res[3] - p2[b]);
        cur--;
    }
    printf("%lld\n", res[1] + mx);
}