Problem Description
Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == 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 ≤ 1,000,000,000 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p,a;如果a的p次方对p取余等于a,并且p不是素数,则输出“yes”,否则输出“no”.
这里用到快速幂求余技巧
#include <iostream> #include <stdio.h> using namespace std; bool isprime(long long n){ for (long long i = 2; i*i <= n; i++){ if (n%i == 0) return false; } return true; } long long qmod(long long a, long long r, long long m){ long long res = 1; while (r){ if (r & 1) res = res*a%m; a = a*a%m; r >>= 1; } return res; } int main(){ long long p, a; while (scanf("%I64d%I64d", &p, &a) && p&&a){ if (!isprime(p) && qmod(a, p, p) == a) printf("yes "); else printf("no "); } return 0; }