Primitive Roots
We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7. Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p.
Input
Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.
Output
For each p, print a single number that gives the number of primitive roots in a single line.
Sample Input
23
31
79
Sample Output
10
8
24
Source
贾怡@pku
题目类型:欧拉函数
算法分析:由于素数p有phi(p - 1)个原根,所以直接计算在p - 1处的欧拉函数值即可
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 7; const int maxn = 10000000 + 66; long long GetEuler (long long val) { long long ans = val; for (long long i = 2; i * i <= val; i++) { if (val % i == 0) { ans -= ans / i; while (val % i == 0) { val /= i; } } } if (val > 1) ans -= ans / val; return ans; } int main() { // ifstream cin ("aaa.txt"); long long val; while (cin >> val) { cout << GetEuler (val - 1) << endl; } return 0; } |