ap = a (mod p)

最新推荐文章于 2021-07-11 11:19:31 发布

idealistic

最新推荐文章于 2021-07-11 11:19:31 发布

阅读量1.1k

点赞数

CC 4.0 BY-SA版权

分类专栏：快速幂编程语言

本文链接：https://blog.youkuaiyun.com/idealistic/article/details/52004226

编程语言同时被 2 个专栏收录

307 篇文章

订阅专栏

快速幂

10 篇文章

订阅专栏

本文介绍了一道编程题目，该题目要求通过费马小定理来判断一个数是否为特定基数下的伪素数。文章提供了一个C++实现的示例代码，其中包括了快速幂算法模板及质数判断函数。

B - B

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

Sample Output

no
no
yes
no
yes
yes

如果理解题意就很简单了。。

题解：先判断p是不是质数，如果是输出no然后判断a的p次方对p取模是否等于a

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
#include<stdio.h>
#include<math.h>
int s(long long a)
{
	if(a==2)
	return 1;
	for(int i=2;i*i<=a;i++)
	if(a%i==0)
	return 0;
	return 1;
}
long long f(long long a,long long b,long long c)//快速幂模板
{
	long long t=1%c;
	a=a%c;
	while(b>0)
	{
		if(b%2==1)
		t=t*a%c;
		b=b/2;
		a=a*a%c;
	}
	return t;
}
int main()
{
__int64 a,p;
	while(~scanf("%I64d%I64d",&p,&a)&&!(a==0&&p==0))
	{
		if(s(p))
		printf("no\n");
		else
		{
			if(f(a,p,p)==a)
    	    printf("yes\n");
	       else
	       printf("no\n");	
		}
	
	}
	return 0;
}