Pseudoprime numbers poj3641(快速幂+素数判定)
2014-08-08 20:47
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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
Sample Output
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 题意:给一个 p 和 一个 a,如果这个p 本身就是一个素数,就输出 no,如果不是素数,那么计算 ( a ^ p) % p 如果结果等于 a 那么输出 yes 否则输出 no#include<stdio.h> #include<iostream> #include<memory.h> #include<algorithm> #include<math.h> #define N 1000000000 using namespace std; typedef long long ll; int pri(ll n) { int i; if(n==2) return 0; for(i=2;i*i<=n;i++) if(n%i==0) return 0; return 1; } ll pow_mod(ll x,ll n,ll mod) { ll ans=1; while(n) { if(n&1) ans=ans*x%mod; x=x*x%mod; n>>=1; } return ans; } int main() { ll p,a; while(scanf("%I64d%I64d",&p,&a),p&&a) { if(pri(p)==1) { printf("no\n"); continue; } ll ans=pow_mod(a,p,p); if(ans==a) printf("yes\n"); else printf("no\n"); } return 0; }
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