标签:poj3641
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6644 | Accepted: 2696 |
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
#include <stdio.h> #include <string.h> #include <math.h> typedef long long LL; bool isPrime(int n) { if(n < 2) return false; int t = sqrt((double)n); for(int i = 2; i <= t; ++i) if(n % i == 0) return false; return true; } LL mod_power(LL x, LL n, LL mod) { LL ret = 1; for( ; n > 0; n >>= 1) { if(n & 1) ret = ret * x % mod; x = x * x % mod; } return ret; } int main() { LL p, a; while(scanf("%lld%lld", &p, &a), p | a) { if(isPrime(p)) printf("no\n"); else printf(mod_power(a, p, p) == a ? "yes\n" : "no\n"); } return 0; }
POJ3641 Pseudoprime numbers 【快速幂】
标签:poj3641
原文地址:http://blog.csdn.net/chang_mu/article/details/41173473