The answer is the number of $$$1$$$s modulo $$$2$$$.

We can get that by adding '-' before the $$$\text{2nd}, \text{4th}, \cdots, 2k\text{-th}$$$ $$$1$$$, and '+' before the $$$\text{3rd}, \text{5th}, \cdots, 2k+1\text{-th}$$$ $$$1$$$.

#include <bits/stdc++.h>
using namespace std;

char c[1005];
int main() {
  int t;
  scanf("%d", &t);
  int n;
  while (t--) {
    scanf("%d", &n);
    scanf("%s", c + 1);
    int u = 0;
    for (int i = 1; i <= n; ++i) {
      bool fl = (c[i] == '1') && u;
      u ^= (c[i] - '0');
      if (i != 1) putchar(fl ? '-' : '+');
    }
    putchar('\n');
  }
}

If for some $$$i$$$, $$$a_i > \left\lceil\frac{n}{k}\right\rceil$$$. We can divide $$$n$$$ cells into $$$\left\lceil\frac{n}{k}\right\rceil$$$ segments which are not longer than $$$k$$$. According to the pigeonhole principle, at least one contains two cells with the color $$$i$$$.

#include <bits/stdc++.h>
using namespace std;

int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    int fl = 0;
    for (int i = 1; i <= m; ++i) {
      int a;
      scanf("%d", &a);
      if (a == (n + k - 1) / k) ++fl;
      if (a > (n + k - 1) / k) fl = 1 << 30;
    }
    puts(fl <= (n - 1) % k + 1 ? "YES" : "NO");
  }
}