快速幂：快速幂就是所求的幂次方过大，导致代码所用的时间超限。
如：求2^3,3的二进制是11，(n&1)判断次方数的二进制是否为1，n>>1,向右进位1：
代码：
k=1,t=n;
while(n)
{
if(n&1)//判断n的最后一位二进制不为0
{
k=k*m;
}
n=n>>1;
m=m*m;
}
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 ≤ 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
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
Sample Output
no
no
yes
no
yes
yes
解题思路：这个题理解起来就是两个函数去判断，对应输出yes/no,首先判断这个数是否为素数，然后再判断(a^p)%p==a就可以了，不过这个幂次方就是需要快速幂。
程序代码：
#include<stdio.h>
#include<math.h>
int fn(long long n)
{
long long i,j,k;
k=sqrt(n);
for(i=2;i<=k;i++)
{
if(n%i==0)
return 0;
}
return 1;
}
int f(long long n,long long m)
{
long long k,a,t;
k=1,t=n;
while(n)
{
if(n&1)
{
k=(k*m)%t;
}
n=n>>1;
m=(m*m)%t;
}
return k;
}
int main()
{
long long i,j,k,m,n;
while(scanf("%lld%lld",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
if(fn(n)==1)
printf("no\n");
else
{
k=f(n,m);
if(k==m)
printf("yes\n");
else
printf("no\n");
}
}
return 0;
}