Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10620 | Accepted: 4578 |
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:输入两个数,判断公式 a^p=a(mod p)是否成立,即a的p次方对p求模是否等于a。只有当p是非素数时且等式成立时,才输出yes。其他情况输出no。
题解:利用快速幂算法,简单的素数判定。
代码如下:
#include<cstdio>
typedef long long ll;
bool Prime_judge(ll q)
{
for(ll i=2;i*i<=q;i++)
{
if(q%i==0) return false;
}
return true;
}
ll p,a;
ll mod_pow(ll x,ll n){
ll res=1;
while(n>0){
if(n&1) res=res*x%p;
x=x*x%p;
n>>=1;
}
return res;
}
int main()
{
while(scanf("%lld%lld",&p,&a)){
if(p==0&&a==0) break;
if(Prime_judge(p)){
printf("no\n");
}else{
ll ans=mod_pow(a,p);
ans=ans%p;
if(ans==a) printf("yes\n");
else printf("no\n");
}
}
return 0;
}