此题与象棋1类似仅需把1中的2换成k即可。

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

using i64 = long long;

i64 power(i64 a, i64 n, i64 p)
{
	i64 ans = 1;
	while (n)
	{
		if (n & 1)  (ans *= a) %= p;
		(a *= a) %= p;
		n >>= 1;
	}
	return ans;
}

i64 inv(i64 x)
{
	static const i64 mod = 998244353;
	return power(x, mod - 2, mod);
}

void solve()
{
	static const i64 mod = 998244353;
	i64 a, b, k;
	cin >> a >> b >> k;
	i64 p = (a % mod) * inv(b % mod) % mod;
	i64 y = 2 * p - 1;
	i64 ans = k * inv(y % mod);
	cout << ans % mod << "\n";
}

int main()
{
	int T = 1; cin >> T;
	while (T--) solve();	
	return 0;
}