P4449 于神之怒加强版

$\color{#0066ff}{ 题目描述 }$

给定n,m,k,计算

$\sum_{i=1}^n \sum_{j=1}^m \mathrm{gcd}(i,j)^k$

对1000000007取模的结果

$\color{#0066ff}{输入格式}$

多组数据。第一行是两个数T,K; 之后的T行，每行两个整数n，m；

$\color{#0066ff}{输出格式}$

K行，每行一个结果

$\color{#0066ff}{输入样例}$

1 2

3 3

$\color{#0066ff}{输出样例}$

$\color{#0066ff}{数据范围与提示}$

T<=2000,1<=N,M,K<=5000000

$\color{#0066ff}{ 题解 }$

就是要求

\[\sum_{i=1}^n \sum_{j=1}^m gcd(i,j)^k
\]

枚举gcd

\[\sum_{d=1}^{min(n,m)}\sum_{i=1}^n \sum_{j=1}^m [gcd(i,j)==d]d^k
\]

把$d^k$提出来，d再除上去，就是一个基本模型了

\[\sum_{d=1}^{min(n,m)}d^k\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} [gcd(i,j)==1]
\]

\[\sum_{d=1}^{min(n,m)}d^k\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} \sum_{k|gcd(i,j)} \mu(k)
\]

\[\sum_{d=1}^{min(n,m)}d^k\sum_{k=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}\mu(k)\sum_{i=1}^{\lfloor\frac{n}{kd}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{kd}\rfloor} 1
\]

后面好像空了。。。

\[\sum_{d=1}^{min(n,m)}d^k\sum_{k=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}\mu(k) * \lfloor\frac{n}{kd}\rfloor * \lfloor\frac{m}{kd}\rfloor
\]

来一发kd换q

\[\sum_{q=1}^{min(n,m)} \lfloor\frac{n}{q}\rfloor * \lfloor\frac{m}{q}\rfloor \sum_{d|q}\mu(\frac q d)*d^k
\]

“额，这怎么处理”

“暴力了解一下”

线性筛出$\mu$ 然后$O(nlogn)$求出$d^k$

之后枚举倍数$O(nlogn)把后面的$\sum$搞出来，数列分块就行了

#include<bits/stdc++.h>

#define LL long long

LL in() {

	char ch; LL x = 0, f = 1;

	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);

	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));

	return x * f;

}

const int mod = 1e9 + 7;

const int maxn = 5e6 + 100;

int k;

int mu[maxn], pri[maxn], tot, mi[maxn];

bool vis[maxn];

LL h[maxn];

LL ksm(LL x, LL y) {

	LL re = 1LL;

	while(y) {

		if(y & 1) re = re * x % mod;

		x = x * x % mod;

		y >>= 1;

	}

	return re;

}

void predoit() {

	mu[1] = 1;

	for(int i = 2; i < maxn; i++) {

		if(!vis[i]) pri[++tot] = i, mu[i] = -1;

		for(int j = 1; j <= tot && (LL)i * pri[j] < maxn; j++) {

			vis[i * pri[j]] = true;

			if(i % pri[j] == 0) break;

			else mu[i * pri[j]] = -mu[i];

		}

	}

	for(int i = 1; i < maxn; i++) mi[i] = ksm(i, k);

	for(int i = 1; i < maxn; i++)

		for(int j = i; j < maxn; j += i)

			(h[j] += (1LL * mu[j / i] * mi[i] % mod)) %= mod;

	for(int i = 2; i < maxn; i++) (h[i] += h[i - 1]) %= mod;

}

LL work(LL n, LL m) {

	LL ans = 0;

	for(LL l = 1, r; l <= std::min(n, m); l = r + 1) {

		r = std::min(n / (n / l), m / (m / l));

		LL tot1 = (n / l) * (m / l) % mod;

		tot1 = (tot1 * ((h[r] - h[l - 1]) % mod + mod) % mod) % mod;

		(ans += tot1) %= mod;

	}

	return ans;

}

int main() {

	int T;

	for(T = in(), k = in(), predoit(); T --> 0;)

		printf("%lld\n", work(in(), in()));

	return 0;

}