大致的思路是通过询问 a,b,c 和 b,c,d（假设 k=3），判断 a 和 d 是否同奇偶，然后用并查集维护元素之间两两的奇偶异同关系。理论复杂度 $O(n)$，但是每个点（包括样例）都是 tle。当然有可能是我第一次写交互题出的格式错误。检查了两天没检查出来，请求帮助。

#include <iostream>
const int N = 1009;
int n, k;
int ds[N << 1];
int res[N];
int seq[N];
int find(int x) { return ds[x]<0 ? x : (ds[x] = find(ds[x])); }
void merge(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return ;
    if (ds[x] > ds[y]) std::swap(x, y);
    ds[x] += ds[y], ds[y] = x;
}
bool judge(int x, int y) { return find(x) == find(y); }
int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n << 1; i++)
        ds[i] = -1;
    for (int i = 1; i <= n; i++) {
        // printf("i = %d\n", i);
        printf("?");
        for (int j = 1; j <= k; j++) {
            int t = i + j - 1;
            if (t > n) t -= n;
            printf(" %d", t);
        }
        puts("");
        scanf("%d", res + i);
        if (i == 1) continue;
        int x = i - 1, y = i + k - 1;
        if (y > n) y -= n;
        // printf("(x, y) = (%d, %d)\n", x, y);
        if (res[i-1] == res[i]) {
            merge(x, y);
            // puts("[merge]: (x, y)");
            merge(x + n, y + n);
            // puts("[merge]: (x+n, y+n)");
        } else {
            merge(x, y + n);
            // puts("[merge]: (x, y+n)");
            merge(x + n, y);
            // puts("[merge]: (x+n, y)");
        }
    }
    // puts("Ridiculous!");
    for (int i = 2; i <= n; i++)
        if (judge(1, i) == false)
            seq[i] = 1;
    int t = 0;
    for (int i = 1; i <= k; i++)
        t += seq[i];
    if (res[1] != (t & 1)) t = 1;
    else t = 0;
    printf("!");
    for (int i = 1; i <= n; i++)
        printf(" %d", seq[i] ^ t);
    puts("");
    return 0;
}

提交记录：

https://atcoder.jp/contests/abc313/submissions/44395809

https://atcoder.jp/contests/abc313/submissions/44371539

https://atcoder.jp/contests/abc313/submissions/44371418

https://atcoder.jp/contests/abc313/submissions/44371408