HAOI2011 Problem b

题意

给出 \(a,b,c,d,k\) ，求 \(\sum_{i=a}^b\sum_{j=c}^d[gcd(i,j)==k]\) 。

传送

Luogu

BZOJ

分析

假设 \(b\le d\)

将 \(k\) 提出来

\[\sum_{i=a}^{\lfloor\frac{b}{k}\rfloor}\sum_{j=c}^{\lfloor\frac{d}{k}\rfloor}[gcd(i,j)==1] \]

将 \(gcd(i,j)==1\) 替换为 \(\epsilon(gcd(i,j))\)

\[\sum_{i=a}^{\lfloor\frac{b}{k}\rfloor}\sum_{j=c}^{\lfloor\frac{d}{k}\rfloor}\epsilon(gcd(i,j)) \]

再将 \(\epsilon\) 函数用 \(\mu\) 表示

\[\sum_{i=a}^{\lfloor\frac{b}{k}\rfloor}\sum_{j=c}^{\lfloor\frac{d}{k}\rfloor}\sum_{d|gcd(i,j)}\mu(d) \]

交换求和顺序，改为枚举 \(d\)

\[\sum_{d=1}^{\lfloor\frac{b}{k}\rfloor}\mu(d)\sum_{i=1}^{\lfloor\frac{b}{k}\rfloor}d|i\sum_{j=1}^{\lfloor\frac{d}{k}\rfloor}d|j \]

\(\sum_{i=1}^{\lfloor\frac{b}{k}\rfloor}d|i\) 即 \({\lfloor\frac{b}{k}\rfloor}\) 以内 \(i\) 的约数个数和，等于 \(\lfloor\frac{b}{kd}\rfloor\) ，同理， \(\sum_{j=1}^{\lfloor\frac{d}{k}\rfloor}d|j\) 等同于 \(\lfloor\frac{d}{kd}\rfloor\) 。

则式子为

\[\sum_{d=1}^{\lfloor\frac{b}{k}\rfloor}\mu(d)\lfloor\frac{b}{kd}\rfloor\lfloor\frac{d}{kd}\rfloor \]

于是我们便可以用整除分块求解了。

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 500005
#define il inline
#define re register
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;

template <typename T> inline void read(T &x) {
	T f = 1; x = 0; char c;
    for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    x *= f;
}

int a, b, c, d, k;
int mu[N], prime[N];
bool vis[N];

void sieve() {
	int cnt = 0;
	mu[1] = 1;
	for (int i = 2; i <= N; ++i) {
		if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
		for (int j = 1; j <= cnt && i * prime[j] <= N; ++j) {
			vis[i*prime[j]] = 1;
			if (i % prime[j] == 0) {
				mu[i*prime[j]] = 0;
				break;
			}
			mu[i*prime[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= N; ++i) mu[i] += mu[i-1];
}

int f(int n, int m) {
	int sum = 0;
	for (int i = 1, j; i <= min(n, m); i = j + 1) {
		j = min(n / (n / i), m / (m / i));
		sum += (mu[j] - mu[i-1]) * (n / i) * (m / i);
	}
	return sum;
}

int main() {
	int n;
	read(n);
	sieve();
	while (n--) {
		read(a), read(b), read(c), read(d), read(k);
		printf("%d\n", f(b / k, d / k) - f(b / k, (c - 1) / k) - f((a - 1) / k, d / k) + f((a - 1) / k, (c - 1) / k));
	}
	return 0;
}