题面
求\(\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}gcd(i, j)\)
n<=4000000,数据组数T<=100
答案保证在64位带符号整数范围内（long long就好）
Sol
之前做了一道假的
先不管i，j是否有序，我们就求\(\sum_{i=1}^{n}\sum_{j=1}^{n}gcd(i, j)\)
最后\(ans=(ans - (n + 1) * n / 2) / 2\)即可
推导
\(ans=\sum_{d=1}^{n}\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\mu(i)*\lfloor\frac{n}{i*d}\rfloor^2\)
\(用k替换i*d，ans=\sum_{k=1}^{n}\lfloor\frac{n}{k}\rfloor^2\sum_{d|k}\mu(\frac{k}{d})d\)
\(\sum_{d|k}\mu(\frac{k}{d})d\)是积性函数，线性筛即可
加上数论分块
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Zsydalao 666
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e6 + 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[_], num;
ll f[_];
bool isprime[_];
IL void Prepare(){
isprime[1] = 1; f[1] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]) prime[++num] = i, f[i] = i - 1;
for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]) f[i * prime[j]] = f[i] * f[prime[j]];
else{ f[i * prime[j]] = f[i] * prime[j]; break; }
}
}
for(RG int i = 2; i < _; ++i) f[i] += f[i - 1];
}
int main(RG int argc, RG char *argv[]){
Prepare();
while(Zsydalao == 666){
RG ll n = Read(), ans = 0;
if(!n) break;
for(RG ll k = 1, j; k <= n; k = j + 1){
j = n / (n / k);
ans += (n / k) * (n / k) * (f[j] - f[k - 1]);
}
printf("%lld\n", (ans - n * (n + 1) / 2) / 2);
}
return 0;
}