题面
Sol
傻逼题
(原式=sum_{p质数}^{n}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}[gcd(i, j)==1])
(设f(i) = sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}[gcd(i, j)==1])
(g(i) = sum_{i|d} f(d) = lfloorfrac{lfloorfrac{n}{d}
floor}{i}
floorlfloorfrac{lfloorfrac{m}{d}
floor}{j}
floor)
质数就把质数的bool求前缀和就好了，数论分块
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1);
IL ll Read(){
char c = '%'; ll x = 0, z = 1;
for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x * z;
}
int prime[_], mu[_], num, s[_];
bool isprime[_];
IL void Prepare(){
isprime[1] = 1; mu[1] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]) prime[++num] = i, mu[i] = -1;
for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]) mu[i * prime[j]] = -mu[i];
else{ mu[i * prime[j]] = 0; break; }
}
mu[i] += mu[i - 1]; s[i] = s[i - 1] + (!isprime[i]);
}
}
IL ll Calc(RG ll n){
RG ll f = 0;
for(RG ll i = 1, j; i <= n; i = j + 1) j = n / (n / i), f += 1LL * (mu[j] - mu[i - 1]) * (n / i) * (n / i);
return f;
}
int main(RG int argc, RG char *argv[]){
Prepare();
RG int n = Read(); RG ll ans = 0;
for(RG ll d = 1, j; d <= n; d = j + 1) j = n / (n / d), ans += 1LL * (s[j] - s[d - 1]) * Calc(n / d);
printf("%lld
", ans);
return 0;
}