C++ 解法, 执行用时: 3149ms, 内存消耗: 218760K, 提交时间: 2021-11-19 21:55:42

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

typedef long long s64;
const int N = 1e8 + 5, D = 998244353;
s64 pow_mod(s64 x, int y = D - 2) {
	s64 ans = x;
	--y;
	while (y) {
		if (y & 1) ans = ans * x % D;
		x = x * x % D;
		y >>= 1;
	}
	return ans;
}
int n;
char miu[N];
bool mark[N];
int pr[N / 10], pcnt;
void init(int n) {
	miu[1] = 1;
	for (int i = 2; i <= n; ++i) {
		if (!mark[i]) {
			pr[++pcnt] = i;
			miu[i] = -1;
		}
		for (int j = 1; j <= pcnt; ++j) {
			int p = pr[j];
			if (i * p > n) break;
			mark[i * p] = 1;
			if (i % p == 0) break;
			miu[i * p] = -miu[i];
		}
	}
}
int q[N];
s64 F(int n, int x, int y, int a, int b) {
	if (y > x) y = x = 1;
	int t = 0;
	int i = 1;
	while (i <= n) {
		q[++t] = i;
		i = n / (n / i) + 1;
	}
	q[t + 1] = n + 1;
	int j = 0;
	s64 ans = 0;
	s64 w2 = 0, s_i = 0;
	i = 0;
	for (int h = t; h; --h) {
		int d = q[h];
		int m = n / d;
		s64 w1 = 0;
		for (int i = d; i < q[h + 1]; ++i)
			if (miu[i] == 1)
				w1 += pow_mod(i, a + b);
			else if (miu[i] == -1)
				w1 -= pow_mod(i, a + b);
		w1 %= D;
		while (j < m) {
			++j;
			while ((s64)(i + 1) * x < (s64)y * j) {
				++i;
				(s_i += pow_mod(i, a)) %= D;
			}
			(w2 += pow_mod(j, b) * s_i) %= D;
		}
		(ans += w1 * w2) %= D;
	}
	return ans;
}

int main() {
#ifdef kcz
	freopen("1.in", "r", stdin); //freopen("1.out","w",stdout);
#endif
	int x, y, a, b;
	cin >> n >> y >> x >> a >> b;
	int d = __gcd(x, y);
	x /= d;
	y /= d;
	init(n);
	s64 ans = F(n, x, y, a, b);
	if (y > x) {
		ans = (ans + 1 + F(n, 1, 1, b, a) - F(n, y, x, b, a)) % D;
	} else if (x <= n && y <= n)
		(ans += pow_mod(y, a) * pow_mod(x, b)) %= D;
	cout << (ans % D + D) % D << endl;
}