\(\text{Solution}\)

發現題目就是求 \(\sum[\prod_{i=1}^k x_i \le n]\)
\(k \le 10^9\) 太可怕了
然而發現如果限定 \(x_i > 1\) 那麼 \(i \le \log n\)
於是我們可以愉快地統計了
設 \(f_i(n)\) 表示將 \(n\) 分成 \(i\) 份使 \(x_i > 1\) 的方案數
那麼 \(f_i(n)=\sum_{d|n}f_{i-1}(\frac n d)-f_{i-1}(n)\)
那個減號就是減去 \(d=1\) 時的情況
先不考慮減法，發現它的轉移就是 \(Dirichlet\) 字首和
於是處理 \(f\)

\(\text{Code}\)

#include <cstdio>
#include <iostream>
#define re register
#define LL long long
using namespace std;

const int N = 5e5 + 5, P = 998244353;
int mxR, mxK, q, log[N], tot, pr[N], vis[N];
LL f[20][N], inv[20];

inline void read(int &x) 
{
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) f = (ch == '-' ? -1 : f), ch = getchar();
	while (isdigit(ch)) x = (x<<3) + (x<<1) + (ch^48), ch = getchar();
	x *= f;
}
void sieve(int n)
{
	for(re int i = 2; i <= n; i++)
	{
		if (!vis[i]) pr[++tot] = i;
		for(re int j = 1; j <= tot && pr[j] * i <= n; j++)
		{
			vis[pr[j] * i] = 1;
			if (i % pr[j] == 0) break;
		}
	}
}
LL Add(LL x, LL y)
{
	x += y;
	if (x > P) x -= P;
	return x;
}

int main()
{
	freopen("easy.in", "r", stdin), freopen("easy.out", "w", stdout);
	read(mxR), read(mxK), read(q);
	sieve(mxR);
	for(re int i = 2; i <= mxR; i++) log[i] = log[i >> 1] + 1, f[1][i] = 1;
	int lg = log[mxR];
	for(re int i = 2; i <= lg; i++)
	{
		for(re int j = 1; j <= mxR; j++) f[i][j] = f[i - 1][j];
		for(re int j = 1; j <= tot; j++)
			for(re int k = 1; k * pr[j] <= mxR; k++)
				f[i][k * pr[j]] = Add(f[i][k * pr[j]], f[i][k]);
		for(re int j = 1; j <= mxR; j++) f[i][j] = Add(f[i][j], P - f[i - 1][j]);
	}
	for(re int i = 1; i <= lg; i++)
		for(re int j = 1; j <= mxR; j++) f[i][j] = Add(f[i][j], f[i][j - 1]);
	inv[1] = 1;
	for(re int i = 2; i <= lg; i++) inv[i] = (P - P / i) * inv[P % i] % P;
	for(int l, r, k; q; q--)
	{
		read(l), read(r), read(k);
		LL ans = 0, c = 1;
		for(re int i = 1; i <= log[r]; i++)
		{
			c = c * (k - i + 1) % P * inv[i] % P;
			ans = Add(ans, (f[i][r] - f[i][l - 1] + P) * c % P);
		}
		printf("%lld\n", ans + (l <= 1));
	}
}