2090A_Treasur_Hunt.cpp

/*                                                         A. Treasure Hunt
                                                                                                time limit per test 1 second
                                                                                                memory limit per test 256 megabytes
Little B and his friend Little K found a treasure map, and now they just need to dig up the treasure, which is buried at a depth of a.5 meters.
They take turns digging. On the first day, Little B digs; on the second day, Little K.
After each day, they switch. Little B digs exactly x meters of soil each day, while Little K digs y meters.
They became curious about who will dig up the treasure first, meaning during whose day the total dug depth will exceed a.5.
But they are too busy digging, so help them and tell who will dig up the treasure!

Input
The first line contains an integer t (1≤t≤1000) — the number of test cases. The description of the test cases follows.
In a single line of each test case, three integers x,y,a are given (1≤x,y,a≤109).
Output
For each test case, output "NO" if Little B digs it up first; otherwise, output "YES". You can output the answer in any case.
Example

Input:
3
1 2 4
2 1 4
2 2 1
Output:
YES
NO
NO

Note:
In the first test case, on the first day they will dig 1 meter; on the second day 1+2=3 meters in total; on the third day 1+2+1=4
 meters; and on the fourth day they will dig 6 meters. Thus, the treasure will be dug up first by Little K.

In the second test case, on the first day they will dig 2 meters; on the second day 2+1=3 meters in total; on the third day 2+1+2=5
meters, meaning it was dug up first by Little B.
                                                                                                                                              */

#include <iostream>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        long long x, y, a;
        cin >> x >> y >> a;

        long long depth = 0;
        long long total = 0;
        long long full_cycles = (a / (x + y));
        total = full_cycles * (x + y);

        // After full cycles, check who digs last
        if (total + x > a + 0.5) {
            cout << "NO" << endl;  // Little B digs it up
        } else if (total + x + y > a + 0.5) {
            cout << "YES" << endl;  // Little K digs it up
        } else {
            // If still not enough, another full cycle — B starts again
            cout << "NO" << endl;
        }
    }
    return 0;
}