POJ 1284:Primitive Roots 求原根的数量
2024-10-08 15:35:00
Primitive Roots
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 3381 | Accepted: 1980 |
Description
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.
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
一个数m的1次方,2次方,3次方到n-1次方 mod n 得到的数值各不相同,就说m是n的原根。
一个数是数n的原根就必然与n-1互质,所以求n的原根的数量即是求欧拉函数n-1。
代码:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; long long euler(long long n)
{
long long res = n, a = n;
for (long long i = 2; i*i <= a; i++)
{
if (a%i == 0)
{
res = res / i*(i - 1);
while (a%i == 0)a /= i;
}
}
if (a > 1)res = res / a*(a - 1);
return res;
} int main()
{
long long n;
while (cin >> n)
{
cout << euler(n-1) << endl;
}
return 0;
}
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