[POJ 3641] Pseudoprime numbers【快速幂】
Problem:
| Time Limit: 1000MS | Memory Limit: 65536K |
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
Source
Waterloo Local Contest, 2007.9.23
Solution:
又一道大水题。
O(sqrt(n)) 素数判定 + 快速幂模板。
Code: O(Tsqrt(p)) [160K, 16MS]
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll p, a;
inline bool isPrime(ll num){
for(register ll i = 2; i * i <= num; i++) if(num % i == 0) return 0;
return 1;
}
inline ll fastpow(ll bas, ll ex, ll md){
ll iden = bas, res = 1;
while(ex){
if(ex & 1) res = res * iden % md;
ex >>= 1, iden = iden * iden % md;
}
return res;
}
int main(){
while(scanf("%lld%lld", &p, &a) != EOF && p > 2){
if(!isPrime(p) && fastpow(a, p, p) == a) puts("yes");
else puts("no");
}
return 0;
}
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