P1447 [NOI2010]能量采集

gate

一个植物 ((x,y)) 与((0,0))上的连线共有 (gcd(x,y))个点，再减去两个端点，
能量损失即为(2 imes(gcd(x,y)-2)+1 = 2 imes gcd(x,y)+1)

所以题目要求的即：

(sumlimits_{i=1}^nsumlimits_{j=1}^m(2 imes gcd(i,j)-1))

(=2 imessumlimits_{i=1}^nsumlimits_{j=1}^mgcd(i,j) - n imes m)

设(f=sumlimits_{i=1}^nsumlimits_{j=1}^mgcd(i,j))

(=sumlimits_{d=1}^ndsumlimits_{i=1}^nsumlimits_{j=1}^m[gcd(i,j)=d])

(=sumlimits_{d=1}^ndsumlimits_{i=1}^{frac{n}{d}}sumlimits_{j=1}^frac{m}{d}sumlimits_{k|i,k|j}mu(k))

(=sumlimits_{d=1}^ndsumlimits_{k=1}^frac{n}{d}mu(k)dfrac{n}{kd} imesdfrac{m}{kd})

设 (T = kd)

(=sumlimits_{T=1}^ndfrac{n}{T} imesdfrac{m}{T}sumlimits_{k|T}mu(k) imesdfrac{T}{k})

根据(id imesmu=varphi)

(=sumlimits_{T=1}^ndfrac{n}{T} imesdfrac{m}{T} imesvarphi(T))

原式即(=2 imessumlimits_{T=1}^nlfloordfrac{n}{T} floor imeslfloordfrac{m}{T} floor imesvarphi(T) - n imes m)

时间复杂度(O(n))

code

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define MogeKo qwq
using namespace std;

const int maxn = 1e5+10;
const int N = 1e5;

int phi[maxn],prime[maxn],cnt;
long long n,m,sum[maxn];
bool vis[maxn];

void Phi() {
	sum[1] = phi[1] = 1;
	for(int i = 2; i <= N; i++) {
		if(!vis[i]) {
			phi[i] = i-1;
			prime[++cnt] = i;
		}
		for(int j = 1; j <= cnt && i*prime[j] <= N; j++) {
			vis[i*prime[j]] = true;
			if(i % prime[j])
				phi[i*prime[j]] = phi[i] * (prime[j]-1);
			else {
				phi[i*prime[j]] = phi[i] * prime[j];
				break;
			}
		}
		sum[i] = sum[i-1] + phi[i];
	}
}

long long f(long long n,long long m) {
	long long ans = 0;
	if(n > m) swap(n,m);
	for(int i = 1,r; i <= n; i = r+1) {
		r = min(n/(n/i),m/(m/i));
		ans += (n/i)*(m/i)*(sum[r]-sum[i-1]);
	}
	return 2 * ans - n*m;
}

int main() {
	scanf("%lld%lld",&n,&m);
	Phi();
	printf("%lld",f(n,m));
	return 0;
}