GCJ 2019 Round 1A Problem 2 "Golf Gophers" solution with N = 6

golf_gophers.cpp

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;

int T, N, M, sieve[1000005];
int q[6] = { 17, 9, 7, 11, 13, 16 };
int res[6];

void do_q(int j) {
	for (int i = 0; i < 18; i++) {
		if (i == 0) printf("%d", q[j]);
		else printf(" %d", q[j]);
	} printf("\n"); fflush(stdout);
    for (int i = 0; i < 18; i++) {
		int a; scanf("%d", &a);
		res[j] += a;
	}
	res[j] %= q[j];
}

int main() {
	scanf("%d%d%d", &T, &N, &M);
	for (int t = 1; t <= T; t++) {
		memset(res, 0, sizeof(res));
		for (int i = 0; i < 6; i++) {
			do_q(i);
		}
		// Regular for loop brute force because i cbs implementing CRT
		int found = 0;
		for (int i = res[0]; !found && i <= M; i += q[0]) {
			int cooked = 0;
			for (int j = 1; j < 6; j++) if (i % q[j] != res[j]) { cooked = 1; break; }
			if (!cooked) {
				printf("%d\n", i); fflush(stdout);
				found = 1;
			}
		}
		int verdict;
		scanf("%d", &verdict);
		if (verdict == -1) return 0;
	}
	return 0;
}