POJ 3641 Pseudoprime numbers
Pseudoprime numbers
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4864 | Accepted: 1872 |
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
//给你一个数 p和一个数 a, 若p是素数输出no,a的p次方模p等于a输出yes,否则输出no
#include <iostream> #include <stdio.h> #include <cmath> using namespace std; bool is_prime(int n) { int i,m=sqrt(double(n)); for(i=2;i<=m;i++) if(n%i==0) return 0; return 1; } int main() { int q,a,tq; __int64 ta,i; while(scanf("%d%d",&q,&a),q||a) { if(is_prime(q)) { printf("no\n");continue; } tq=q; i=a; for(ta=1;tq>0;tq=tq>>1,i=(i*i)%q) { if(tq&1) ta=(ta*i)%q; } if(ta==a) printf("yes\n"); else printf("no\n"); } return 0; }