Pseudoprime numbers
时间： 1ms 内存：128M
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：
输出：
示例输入：
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
示例输出：
no
no
yes
no
yes
yes
提示：
参考答案（内存最优[752]）：
#include <stdio.h>
int isprime(long long p)
{
long long i;
for(i = 2; i*i <= p; i++)
{
if(p%i == 0) return 0;
}
return 1;
}
int main()
{
long long p, a, ap, pp, aa;
while(1)
{
scanf("%lld %lld", &p, &a);
/* scanf("%I64d %I64d", &p, &a);*/
if(p == 0) break;
if(isprime(p))
{
puts("no");
continue;
}
ap = 1;
pp = p;
aa = a;
while(pp)
{
if(pp&1) ap*= aa;
ap %= p;
aa = (aa*aa)%p;
pp >>= 1;
}
/* printf("%lld %lld\n", ap%p, a);*/
if(ap%p == a) puts("yes");
else puts("no");
}
}
参考答案（时间最优[0]）：
#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("yes\n");
else printf("no\n");
}
return 0;
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