题目描述
Fermat's theorem states that for any prime number p and for any integer a > 1, ap == a (mod p). That is, if we raise a to the pth 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 ≤ 1,000,000,000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.
Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a. For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".
输入
暂无
输出
暂无
样例输入
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
样例输出
no
no
yes
no
yes
yes
参考代码
#include<stdio.h>
int pp(int a)
{
int i;
for (i=2; i*i<=a; i++)
if(a%i==0)return 0;
return 1;
}
int mod(long x,long n,long m)
{
long a,r=1;
a=x;
while(n)
{
if(n&1)r=r*x%m;
x=x*x%m;
n>>=1;
}
return r==a?1:0;
}
int main()
{
int a,p;
while(~scanf("%d%d",&p,&a)&&a||p)
{
if(!pp(p)&&mod(a,p,p))printf("yesn");
else printf("non");
}
return 0;
}
解析
暂无
