【BZOJ3930】【CQOI2015】选数

【题目链接】

• 点击打开链接
  

【思路要点】

• 显然问题可以转化为\(K=1\)的形式。
• 那么，我们实际上要求\(\sum_{i_1,i_2,...,i_N=L}^{R}\epsilon(gcd(i_1,i_2,...,i_N))\)。
• \(=\sum_{i_1,i_2,...,i_N=L}^{R}\sum_{d/i_1,i_2,...,i_N}\mu(d)\)
• \(=\sum_{d=1}^{R}\mu(d)(\lfloor\frac{R}{d}\rfloor-\lfloor\frac{L-1}{d}\rfloor)^N\)
• 其中\((\lfloor\frac{R}{d}\rfloor-\lfloor\frac{L-1}{d}\rfloor)^N\)只有\(O(\sqrt{L}+\sqrt{R})\)种取值，而\(\mu(d)\)的前缀和可以通过杜教筛求得。
• 时间复杂度\(O(L^{\frac{2}{3}}+R^{\frac{2}{3}}+(\sqrt{L}+\sqrt{R})*LogN)\)。

【代码】

#include
using namespace std;
const int MAXN = 2e6 + 5;
const int P = 1e9 + 7;
template  void chkmax(T &x, T y) {x = max(x, y); }
template  void chkmin(T &x, T y) {x = min(x, y); } 
template  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  void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template  void writeln(T x) {
	write(x);
	puts("");
}
int tot, prime[MAXN], f[MAXN];
int miu[MAXN], s[MAXN];
int sl[MAXN], sr[MAXN];
int n, l, r, k;
void init() {
	miu[1] = s[1] = 1;
	for (int i = 2; i < MAXN; i++) {
		if (f[i] == 0) {
			f[i] = prime[++tot] = i;
			miu[i] = -1;
		}
		s[i] = (s[i - 1] + miu[i] + P) % P;
		for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
			int tmp = prime[j] * i;
			if (tmp >= MAXN) break;
			f[tmp] = prime[j];
			if (prime[j] == f[i]) miu[tmp] = 0;
			else miu[tmp] = -miu[i];
		}
	}
}
int getsl(int n) {
	int m = l / n, ans = 1;
	if (n < MAXN) return s[n];
	else if (sl[m] != -1) return sl[m];
	int now = 2;
	while (now <= n) {
		int nxt = n / (n / now) + 1;
		ans = (ans - 1ll * (nxt - now) * getsl(n / now) % P + P) % P;
		now = nxt;
	}
	return sl[m] = ans;
}
int getsr(int n) {
	int m = r / n, ans = 1;
	if (n < MAXN) return s[n];
	else if (sr[m] != -1) return sr[m];
	int now = 2;
	while (now <= n) {
		int nxt = n / (n / now) + 1;
		ans = (ans - 1ll * (nxt - now) * getsr(n / now) % P + P) % P;
		now = nxt;
	}
	return sr[m] = ans;
}
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;
}
int main() {
	read(n), read(k), read(l), read(r);
	if (l % k == 0) l = l / k - 1;
	else l = l / k; r /= k;
	memset(sl, -1, sizeof(sl));
	memset(sr, -1, sizeof(sr));
	init();
	int now = 1, last = 0, ans = 0;
	while (now <= r) {
		int nxtl, nxtr;
		if (now <= l) nxtl = l / (l / now);
		else nxtl = r;
		nxtr = r / (r / now);
		if (nxtr <= nxtl) {
			ans = (ans + 1ll * (getsr(nxtr) - last + P) * power(r / now - l / now + P, n)) % P;
			last = getsr(nxtr); now = nxtr + 1;
		} else {
			ans = (ans + 1ll * (getsl(nxtl) - last + P) * power(r / now - l / now + P, n)) % P;
			last = getsl(nxtl); now = nxtl + 1;
		}
	}
	writeln(ans);
	return 0;
}