【校内训练2019-04-10】送你一道签到题

【思路要点】
不难看出，计算式中 $i$ 对答案的贡献是一个积性函数。

可以直接用动态规划预处理 $O(p^k)$ 的函数值，再用 $Min25$ 筛计算答案。

时间复杂度 $O(Min25(N)+K^2+K\sqrt{N}+Log^4N)$ 。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int MAXK = 1e3 + 5;
const int MAXE = 45;
const int P = 998244353;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int tot, prime[MAXN], f[MAXN], primek[MAXN], ps[MAXN];
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
ll n, val[MAXN]; int limit, m, k, cnt;
int sum[MAXN], home1[MAXN], home2[MAXN];
int inv[MAXK], s[MAXK][MAXK], func[MAXE];
void init(int n, int Log) {
	for (int i = 2; i <= n; i++) {
		if (f[i] == 0) {
			prime[++tot] = f[i] = i;
			primek[tot] = power(i, k);
			ps[tot] = (ps[tot - 1] + primek[tot]) % P;
		}
		for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
			int tmp = prime[j] * i;
			if (tmp > n) break;
			f[tmp] = prime[j];
		}
	}
	s[0][0] = inv[0] = 1;
	for (int i = 1; i <= k + 1; i++) {
		inv[i] = power(i, P - 2);
		for (int j = 1; j <= i; j++)
			s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % P;
	}
	static int dp[MAXE][MAXE][MAXE];
	static int fac[MAXE], inf[MAXE];
	fac[0] = inf[0] = 1;
	for (int i = 1; i <= Log; i++) {
		fac[i] = 1ll * fac[i - 1] * i % P;
		inf[i] = power(fac[i], P - 2);
	}
	dp[0][0][0] = 1;
	for (int i = 1; i <= Log; i++)
	for (int j = 0; j <= Log; j++)
	for (int k = 0; k <= Log; k++) {
		int res = 0;
		for (int l = 0; l <= j && l * i <= k; l++)
			update(res, 1ll * inf[l] * power(i + 1, l) % P * dp[i - 1][j - l][k - l * i] % P);
		dp[i][j][k] = res;
	}
	func[0] = 1;
	for (int i = 1; i <= Log; i++) {
		int res = 0;
		for (int j = 1; j <= Log; j++) {
			int tmp = dp[Log][j][i];
			for (int k = 1; k <= j; k++)
				tmp = 1ll * tmp * (m - k + 1) % P;
			update(res, tmp);
		}
		func[i] = res;
	}
}
int calc(ll n) {
	n %= P;
	int ans = 0, now = n + 1;
	for (int i = 1; i <= k && i <= n; i++) {
		now = 1ll * now * (n - i + 1) % P;
		update(ans, 1ll * now * inv[i + 1] % P * s[k][i] % P);
	}
	return ans;
}
int solve(ll x, int y) {
	if (x <= 1 || prime[y] > x) return 0;
	int ans = 0, pos = 0;
	if (x <= limit) pos = home1[x];
	else pos = home2[n / x];
	ans = 1ll * func[1] * (sum[pos] - ps[y - 1] + P) % P;
	for (int i = y; i <= tot && 1ll * prime[i] * prime[i] <= x; i++) {
		int coef = primek[i]; ll now = prime[i], nxt = 1ll * prime[i] * prime[i];
		for (int j = 1; nxt <= x; j++, now *= prime[i], nxt *= prime[i], coef = 1ll * coef * primek[i] % P)
			update(ans, (1ll * coef * func[j] % P * solve(x / now, i + 1) + 1ll * coef * primek[i] % P * func[j + 1]) % P);
	}
	return ans;
}
int main() {
	freopen("count.in", "r", stdin);
	freopen("count.out", "w", stdout);
	read(n), read(m), read(k);
	init(limit = sqrt(n) + 1, 40);
	for (ll i = 1, nxt; i <= n; i = nxt + 1) {
		ll tmp = n / i; nxt = n / tmp;
		val[++cnt] = tmp;
		if (tmp <= limit) home1[tmp] = cnt;
		else home2[n / tmp] = cnt;
		sum[cnt] = (calc(tmp) - 1 + P) % P;
	}
	for (int i = 1; i <= tot; i++) {
		for (int j = 1; 1ll * prime[i] * prime[i] <= val[j]; j++) {
			ll tmp = val[j] / prime[i]; int pos;
			if (tmp <= limit) pos = home1[tmp];
			else pos = home2[n / tmp];
			update(sum[j], P - 1ll * primek[i] * (sum[pos] - ps[i - 1] + P) % P);
		}
	}
	writeln((solve(n, 1) + 1) % P);
	return 0;
}
【校内训练2019-04-10】送你一道签到题

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