数论，gcd

gcd(i+j, i-j) = gcd(2*i, i+j)

把i+j看出一个整体，得到范围为[i+1...2*i-1]

再看phi[2i]的意义，即[1...2i-1]上与2*i互质的数的个数

i+j的范围刚好是其一半，又引理可知欧拉函数在这个范围内均分成两半。

所以当i为奇数时，2和i互质，根据积性函数的性质，phi[2i] = phi[2] phi[i] = phi[i]

当i为偶数时，由欧拉函数的性质，phi[2i] = 2phi[i]

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 20000005;int _, n, tot, phi[N], prime[N];ll a[N];bool vis[N];void calc(){    phi[1] = 1;    for(int i = 2; i < N; i ++){        if(!vis[i]){            prime[++tot] = i;            phi[i] = i - 1;        }        for(int j = 1; j <= tot && i * prime[j] < N; j ++){            vis[i * prime[j]] = true;            if(i % prime[j] == 0){                phi[i * prime[j]] = phi[i] * prime[j];                break;            }            phi[i * prime[j]] = phi[i] * (prime[j] - 1);        }    }    for(int i = 1; i < N; i ++){        if(i & 1) a[i] = a[i - 1] + phi[i] / 2;        else a[i] = a[i - 1] + phi[i];    }}int main(){    calc();    for(_ = read(); _; _ --){        n = read();        printf("%lld\n", a[n]);    }    return 0;}