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
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; }
