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

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 
    
     # 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;}