poj 1284 Primitive Roots
2017-04-08 10:28
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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:
给出一个奇素数p(就是不包含2的素数),求出0到p-1有几个原根。
原根的定义:x的1到p-1次方对p取余所得的值的集合,恰好等于1到p-1的集合。
思路:
已知定理我们可知道p的原根个数为φ(p-1)。这里的φ(x)表示的是欧拉函数,求的是小于x的数中与x互质的数的数目。
示例程序
#include <cstdio> using namespace std; int f(int n) { int i,ans; ans=n; for(i=2;n>=i*i;i++) { if(n%i==0) { ans=ans-ans/i; do { n=n/i; }while(n%i==0); } } if(n!=1) { ans=ans-ans/n; } return ans; } int main() { int n; while(scanf("%d",&n)!=EOF) { printf("%d\n",f(n-1)); } return 0; }
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