|
|
| （未显示同一用户的1个中间版本） |
| 第1行： |
第1行： |
| ==求单个欧拉函数==
| | #REDIRECT [[06-数学相关/03-欧拉函数]] |
| | | [[Category:三三文档]] |
| <syntaxhighlight lang="cpp" line>
| |
| int phi(int n)
| |
| {
| |
| int ans = n;
| |
| for (int i = 2; 1LL * i * i <= m; i++)
| |
| if (n % i == 0)
| |
| {
| |
| ans = ans / i * (i - 1);
| |
| while (n % i == 0)
| |
| n /= i;
| |
| }
| |
| if (n > 1)
| |
| ans = ans / n * (n - 1);
| |
| return ans;
| |
| }
| |
| </syntaxhighlight>
| |
| | |
| ==线性筛求欧拉函数==
| |
| | |
| <syntaxhighlight lang="cpp" line>
| |
| const int MAXN = 40000;
| |
| bool p[MAXN + 5];
| |
| int phi[MAXN + 5];
| |
| vector<int> pri;
| |
| // 筛出 1~n 中的每个数是否为质数
| |
| // 顺带求出所有欧拉函数
| |
| void get_primes(int n)
| |
| {
| |
| for (int i = 1; i <= n; i++)
| |
| p[i] = true;
| |
| p[0] = p[1] = false;
| |
| phi[1] = 1;
| |
| for (int i = 2; i <= n; i++)
| |
| {
| |
| if (p[i])
| |
| {
| |
| pri.push_back(i);
| |
| phi[i] = i - 1;
| |
| }
| |
| for (int j = 0; j < pri.size(); j++)
| |
| {
| |
| // i*pri[j]
| |
| if (1LL * i * pri[j] > n)
| |
| break;
| |
| p[i * pri[j]] = false;
| |
| if (i % pri[j] == 0)
| |
| {
| |
| phi[i * pri[j]] = phi[i] * pri[j];
| |
| break;
| |
| }
| |
| else
| |
| phi[i * pri[j]] = phi[i] * (pri[j] - 1);
| |
| }
| |
| }
| |
| }
| |
| </syntaxhighlight>
| |