Poj 3641 Pseudoprime numbers 快速冪取模
阿新 • • 發佈:2018-12-14
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
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
Sample Input
3 2 10 3 341 2 341 3 1105 2 1105 3 0 0
Sample Output
no no yes no yes yes
很基礎的一道快速冪取模的題目....
只需要判斷a^p%p與a%p兩個結果是否相等和判斷p是否為素數即可...
程式碼如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <math.h> using namespace std; typedef long long ll; ll p,a; ll Fast (ll a,ll b) { ll sum=1,mod=b; while (b>0) { if(b&1) { sum=sum*a%mod; } b>>=1; a=a*a%mod; } return sum; } bool Is_pri (ll x) { if(x==1) return false; for (int i=2;i<=sqrt(x);i++) if(x%i==0) return false; return true; } int main() { while (scanf("%lld%lld",&p,&a)!=EOF&&(p||a)) { ll mod1=a%p; ll mod2=Fast(a,p); if(mod2==mod1) { if(Is_pri(p)) printf("no\n"); else printf("yes\n"); } else printf("no\n"); } return 0; }