【LOJ3069】「2019 集训队互测 Day 1」整点计数

【题目链接】

点击打开链接

【思路要点】

给出结论：
定义函数 $\chi(x)\ (x\in\mathbb{N^{+}})$ ，满足 $\chi(x)=\left\{\begin{array}{rcl}1&&{x\equiv 1\ (mod\ 4)}\\-1&&{x\equiv 3\ (mod\ 4)}\\0&&{x\equiv 0\ (mod\ 2)}\end{array} \right.$
令 $x^2=p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_s^{k_s}$ ，其中 $p_1,p_2,p_3,...,p_s$ 均为质数， $k_1,k_2,k_3,...,k_s$ 均为正整数。那么有 $f(x)=4\times\prod_{i=1}^{s}\sum_{j=0}^{k_i}\chi(p_i^j)=4\times\sum_{i\mid x}\chi(i)$

因此 $f(x)$ 除去 $4$ 后为积性函数，使用传统的积性函数求和即可。

关于结论的证明可以参考笔者 $2019$ 年的国家集训队论文。

时间复杂度 $O(Min25(N))$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
const int MAXLOG = 256;
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("");
}
ll n, k, val[MAXN], cnt[MAXN][2];
int m, P, k3, home1[MAXN], home2[MAXN];
int limit, tot, tpow[MAXN], prime[MAXN], pre[MAXN][2];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int power(int x, ll 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;
}
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 = ((cnt[pos][1] - pre[y - 1][1]) + (cnt[pos][0] - pre[y - 1][0]) % P * tpow[3]) % P;
	if (prime[y] == 2) update(ans, 1);
	for (int i = y; i <= tot && 1ll * prime[i] * prime[i] <= x; i++) {
		ll now = prime[i], nxt = 1ll * prime[i] * prime[i];
		for (int j = 1; nxt <= x; j++, now *= prime[i], nxt *= prime[i]) {
			if (prime[i] % 4 == 1) update(ans, (1ll * tpow[2 * j + 1] * solve(x / now, i + 1) + tpow[2 * j + 3]) % P);
			else update(ans, (solve(x / now, i + 1) + 1) % P);
		}
	}
	return ans;
}
void sieve(int n) {
	static int f[MAXN];
	for (int i = 2; i <= n; i++) {
		if (f[i] == 0) prime[++tot] = f[i] = i;
		for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
			int tmp = prime[j] * i;
			if (tmp > n) break;
			f[tmp] = prime[j];
		}
	}
	for (int i = 1; i <= tot; i++) {
		pre[i][0] = pre[i - 1][0];
		pre[i][1] = pre[i - 1][1];
		if (prime[i] % 4 == 1) pre[i][0]++;
		if (prime[i] % 4 == 3) pre[i][1]++;
	}
}
int main() {
	read(n), read(k), read(P);
	sieve(limit = sqrt(n) + 5);
	for (ll i = 1, nxt; i <= n; i = nxt + 1) {
		ll tmp = n / i;
		val[++m] = tmp;
		if (tmp <= limit) home1[tmp] = m;
		else home2[n / tmp] = m;
		nxt = n / val[m];
		cnt[m][0] = (tmp - 1) / 4;
		cnt[m][1] = (tmp + 1) / 4;
	}
	for (int i = 2; i <= tot; i++)
	for (int j = 1; 1ll * prime[i] * prime[i] <= val[j]; j++) {
		ll tmp = val[j] / prime[i]; int pos = 0;
		if (tmp <= limit) pos = home1[tmp];
		else pos = home2[n / tmp];
		if (prime[i] % 4 == 1) {
			cnt[j][0] -= cnt[pos][0] - (pre[i][0] - 1);
			cnt[j][1] -= cnt[pos][1] - pre[i][1];
		} else {
			cnt[j][0] -= cnt[pos][1] - (pre[i][1] - 1);
			cnt[j][1] -= cnt[pos][0] - pre[i][0];
		}
	}
	for (int i = 1; i <= MAXLOG; i++)
		tpow[i] = power(i, k);
	writeln(1ll * tpow[4] * (solve(n, 1) + 1) % P);
	return 0;
}

【LOJ3069】「2019 集训队互测 Day 1」整点计数

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