잉카달력

#include<cstdio>
#include<iostream>
#include<utility>

using namespace std;

typedef pair<int, int> PAIR;
int n;
PAIR ending;
PAIR now;

int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        int x, y;
        cin>>x>>y;
        ending.first = x; ending.second = y;
        int xx,yy;
        cin>>xx>>yy;
        now.first = xx; now.second = yy;
        PAIR comb = make_pair(1,1);
        int answer = 1;
        if(now.first == 1 && now.second == 1)
            answer = 1;
        else {
            while (true) {
                if (comb.first == ending.first && comb.second == ending.second) {
                    answer = -1;
                    break;
                }
                comb.second += 1;
                comb.first += 1;
                answer += 1;
                if (comb.second > ending.second)
                    comb.second = 1;
                if (comb.first > ending.first)
                    comb.first = 1;
                if (comb.first == now.first && comb.second == now.second)
                    break;

            }
        }
        cout<<answer<<endl;
    }
    return 0;

}

 

pair써서 막 여러개해봤는데, 결과는 경우의수가 40000 * 40000 인지라, 너무 커서 시간초과

 

#include<iostream>
#include<utility>

using namespace std;

typedef pair<int, int> PAIR;

int t;

int main() {
	cin >> t;
	for (int al = 0; al < t; al++) {
		int m, n, x, y;
		cin >> m >> n >> x >> y;
		x--;
		y--;
		bool t = false;
		for (int i = x; i < n * m; i += m) {
			if (i % n == y) {
				cout << i + 1 << endl;
				t = true;
				break;
			}
		}
		if (t == false) {
			cout << -1 << endl;
		}
	}
	return 0;
}

조금 건너뛰어가면서 해본결과