gate

一個植物 \((x,y)\) 與\((0,0)\)上的連線共有 \(gcd(x,y)\)個點，再減去兩個端點，
能量損失即為\(2\times(gcd(x,y)-2)+1 = 2\times gcd(x,y)+1\)

所以題目要求的即：

\(\sum\limits_{i=1}^n\sum\limits_{j=1}^m(2\times gcd(i,j)-1)\)

\(=2\times\sum\limits_{i=1}^n\sum\limits_{j=1}^mgcd(i,j) - n\times m\)

設\(f=\sum\limits_{i=1}^n\sum\limits_{j=1}^mgcd(i,j)\)

\(=\sum\limits_{d=1}^nd\sum\limits_{i=1}^n\sum\limits_{j=1}^m[gcd(i,j)=d]\)

\(=\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\frac{n}{d}}\sum\limits_{j=1}^\frac{m}{d}\sum\limits_{k|i,k|j}\mu(k)\)

\(=\sum\limits_{d=1}^nd\sum\limits_{k=1}^\frac{n}{d}\mu(k)\dfrac{n}{kd}\times\dfrac{m}{kd}\)

設 \(T = kd\)

\(=\sum\limits_{T=1}^n\dfrac{n}{T}\times\dfrac{m}{T}\sum\limits_{k|T}\mu(k)\times\dfrac{T}{k}\)

根據\(id\times\mu=\varphi\)

\(=\sum\limits_{T=1}^n\dfrac{n}{T}\times\dfrac{m}{T}\times\varphi(T)\)

原式即\(=2\times\sum\limits_{T=1}^n\lfloor\dfrac{n}{T}\rfloor\times\lfloor\dfrac{m}{T}\rfloor\times\varphi(T) - n\times m\)

時間複雜度\(O(n)\)

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;
}