POJ -3641Pseudoprime numbers（快速幂+同余定理）

Description
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

题意：给你两个数p和a，如果a的p次方对p取余等于a并且p不是素数，输出yes，否则输出no。

思路：因为数据过大,直接写for循环求次方会超时.这里就要用到快速幂这个技巧了.简单说就是降幂,比如说a的b次方,如果b是偶数,该数就可以表示为(a^2)^(b/2),如果b是奇数,可以 另写一个数保存此时的底数然后幂数减1,又可以继续降幂了,知道最后。优化代码。注意__int64输入防止爆int

代码：

#include<cstdio>int quickpow(__int64 n,__int64 m,__int64 mod)//n是底数，m是幂数，mod是取余数{__int64 ans=1,base=n;while(m){if(m&1)//如果幂数是奇数,则将底数保存在ans中,这样幂数-1变成偶数就可以继续降幂了。 base是幂数。{ans=(base*ans)%mod;//保存并更新多幂底数}base=(base*base)%mod;//更新底数m/=2;//更新幂数 }return ans;}int main(){__int64 n,m;__int64 mod;while(scanf("%I64d%I64d",&n,&m)&&n||m){int sum=0;for(int i=2;i*i<n;i++){if(n%i==0)sum++;}if(sum==0)printf("no\n");else{mod=n;if(quickpow(m,n,mod)==m)printf("yes\n");elseprintf("no\n");}}return 0;}