题解 [ABC020D] LCM Rush
题意:给 n,kn,kn,k,求 ∑i=1nlcm(i,k)\sum\limits_{i=1}^n \operatorname{lcm}(i,k)i=1∑nlcm(i,k)。1≤n,k≤1091\leq n,k\leq 10^91≤n,k≤109。
∑i=1nlcm(i,k)=∑i=1nikgcd(i,k)=k∑i=1nigcd(i,k) \begin{aligned} \sum_{i=1}^n \operatorname{lcm}(i,k)&=\sum_{i=1}^n \dfrac{ik}{\gcd(i,k)}\\ &=k\sum_{i=1}^n \dfrac{i}{\gcd(i,k)} \end{aligned} i=1∑nlcm(i,k)=i=1∑ngcd(i,k)ik=ki=1∑ngcd(i,k)i
枚举 g=gcd(i,k)g=\gcd(i,k)g=gcd(i,k),
∑i=1nlcm(i,k)=k∑g∣k∑i=1,(i,kg)=1⌊ng⌋i \begin{aligned} \sum_{i=1}^n \operatorname{lcm}(i,k)&=k\sum_{g|k}\sum_{i=1,(i,\frac kg)=1}^{\lfloor\frac ng\rfloor}i \end{aligned} i=1∑nlcm(i,k)=kg∣k∑i=1,(i,gk)=1∑⌊gn⌋i
现在问题变成了,给定 m,pm, pm,p,求 ∑i=1,(i,p)=1mi\sum\limits_{i=1,(i,p)=1}^mii=1,(i,p)=1∑mi。可以容斥求解。
具体地,枚举所有 q∣pq|pq∣p,qqq 每个质因子次数不大于 111,则
∑i=1,(i,p)=1mi=∑qq×(∑i=1⌊mq⌋i)×valq \sum\limits_{i=1,(i,p)=1}^mi=\sum_q q\times \left(\sum_{i=1}^{\lfloor\frac mq\rfloor}i\right)\times val_q i=1,(i,p)=1∑mi=q∑q×i=1∑⌊qm⌋i×valq
其中当 qqq 的质因子个数为偶数时 valq=1val_q=1valq=1,否则 valq=−1val_q=-1valq=−1。(111 不算质因子)
/*
name: LCM Rush
id: AT_abc020_d
date: 2023/01/25
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll P = 1e9 + 7;
int n, k, top;
ll ans = 0;
int calc(int p){
int sum = 1;
for(int i = 2; i * i <= p; ++ i){
if(p % i == 0){
++ sum;
p /= i;
}
if(p % i == 0){
return 0;
}
}
if(p != 1){
++ sum;
}
return (sum&1) ? 1 : -1;
}
#define adall(k) ((ll)(k) * ((k)+1) / 2)
ll solve(int g){
int p = k / g, m = n / g;
ll cur = 0;
for(int q = 1; q * q <= p; ++ q){
if(p % q == 0){
cur += calc(q) * q * adall((int)m/q);
int r = p/q;
if(r != q){
cur += calc(r) * r * adall((int)m/r);
}
}
}
return cur % P;
}
int main(){
scanf("%d%d", &n, &k);
for(int g = 1; g * g <= k; ++ g){
if(k % g != 0){
continue;
}
ans += solve(g);
if(g * g != k){
ans += solve(k / g);
}
ans %= P;
}
printf("%lld\n", ans * k % P);
return 0;
}