（数论）欧拉函数

1~N中与N互质的数的个数成为欧拉函数，记为$\varphi \left ( N \right ) $
根据容斥原理易得：

\[\varphi \left ( N \right ) = N \times \prod_{质数p\mid N}^{}\left ( 1-\frac{1}{p} \right ) \]

求一个数的欧拉函数时间复杂度与分解质因数相同，为\(O\left ( \sqrt{N} \right )\)

#include <iostream>
using namespace std;

int eulor(int n) {
    int ans = n;
    for (int i = 2; i <= x / i; i++) {
        if (x % i == 0) {
            x /= i;
        }
        ans = ans / (i - 1) * i;
    }
    if (x > 1) ans = ans / (x - 1) * x;
    return ans;
}

int main() {
    cin >> n;
    while (n--) {
        int x;
        cin >> x;
        cout << eulor(x) << endl;
    }
    return 0;
}

接下来考虑求出1~N中所有数的欧拉函数的算法，这里延续了素数筛的思路：
1.由欧拉函数的定义可得\(\varphi \left ( 1 \right ) = 1\),\(\varphi \left ( n \right ) = n - 1\)（当n是质数）
2.考虑每一个合数\(i\times p_{j}\)都会被它的最小质因子\(p_{j}\)筛去,由\(\varphi \left ( n \right )\)的公式可得，当\(p_{j} \mid i\)时，\(\varphi \left ( i\times p_{j} \right ) = \varphi \left ( i \right ) \times p_{j}\),否则\(\varphi \left ( i\times p_{j} \right ) = \varphi \left ( i \right ) \times p_{j} \times \left ( 1 - \frac{1}{p_{j}} \right )=\varphi \left ( i \right ) \times \left ( 1 - p_{j} \right )\)
时间复杂度为\(O\left ( N \right )\)

#include <iostream>
using namespace std;
const int maxn = 1e6 + 10;

int p[maxn], cnt;
int phi[maxn];
bool st[maxn];
int a[maxn];

long long eulor(int n) {
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!st[i]) {
            p[++cnt] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; p[j] * i <= n; j++) {
            st[p[j] * i] = true;
            if (i % p[j] == 0) {
                phi[p[j] * i] = phi[i] * p[j];
                break;
            }
            phi[p[j] * i] = phi[i] * (p[j] - 1);
        }
    }
    long long res = 0;
    for (int i = 1; i <= n; i++)
        res += phi[i];
    return res;
}

int main() {
    int n, t = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        t = max(t, a[i]);
    }
    eulor(t);
    for (int i = 1; i <= n; i++) {
        cout << phi[a[i]] << endl; 
    }
    return 0;
}

例题 Visible point

除了(0,1),(1,0),(1,1)这三个点：
一个点是可见的,当且仅当它的横纵坐标互质（不然横纵坐标一定可以约分，被约分后的那个点挡住，只考虑下半边的话，也就是求满足1<x<y<=N且gcd(x,y)=1的点数，只需要求出phi的前缀和减去phi[1]。
总的答案就是3 + 2 * (sum[N] - 1)

#include <iostream>
using namespace std;
const int maxn = 1e5 + 10;

int p[maxn], cnt;
int phi[maxn];
bool st[maxn];
int t[maxn];
long long sum[maxn];

long long eulor(int n) {
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!st[i]) {
            p[++cnt] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; p[j] <= n / i; j++) {
            st[i * p[j]] = true;
            if (i % p[j] == 0) {
                phi[i * p[j]] = phi[i]  * p[j];
                break;
            }
            else phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
}


int main() {
    int n, mx = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> t[i];
        mx = max(mx,t[i]);
    }
    eulor(mx);
    for (int i = 1; i <= mx; i++) 
        sum[i] = sum[i - 1] + phi[i];
    for (int i = 1; i <= n; i++)
        printf("%d %d %lld\n", i, t[i], 3 + 2 * (sum[t[i]] - 1));
    return 0;
}