euler_prime

/*
 算术基本定理
 n=p1^a1*p2^a2*p3^a3*...*pn^an

 约数个数：(p1+1)*(p2+1)*(p3+1)*..*(pn+1)
 phi(n)=n*(1-1/p1)*(1-1/p2)*...*(1-1/pn)

 欧拉数：1-n中与n互质的数的个数
 phi(n)=n*(1-1/p1)*(1-1/p2)*...*(1-1/pn)
 若p为质数，则phi(n)=p-1
 小于n且与n互质个数的和=euler[n]*n/2
 (a,n)=1 ==>(n-a,n)=1

 n*n的点阵
 ...
 ...
 ...
 ans=2*(phi(1)+phi(2)+..+phi(n))

*/
void euler_prime(int n) {
    euler[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!v[i]) {
            prime[t++] = i;
            euler[i] = i - 1;
        }
        for (int j = 0; j < t && prime[j] * i <= n; j++) {
            v[prime[j] * i] = 1;
            if (i % prime[j] == 0) {
                euler[i * prime[j]] = euler[i] * prime[j];
                break;
            }
            euler[i * prime[j]] = euler[i] * euler[prime[j]];
        }
    }
}

void fenjie(n) {
    euler(sqrt(n));
    for (int i = 0; i < t; i++) {
        if (n % prime[i] == 0) {
            k++;
            a[k] = prime[i];
            while (n % prime[i] == 0) {
                b[k]++;
                n = n / prime[i];
            }
            if (n == 1) break;
        }
    }
    if (n > 1) {
        k++;
        a[k] = n;
        b[k] = 1;
    }
}

int yishugeshu() {
    int res = 1;
    for (int i = 1; i <= k; i++) {
        res = res * (b[i] + 1);
    }
    return res;
}