poj_3641_Pseudoprime numbers
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:费马定理给出a^p=a mod p(p为素数),一些合数也有类似的状况,判断输入p,a
先判断 p是否为素数,后判断是否满足定理
#include<iostream> #include<cstdio> #define LL long long #define N 100000 using namespace std; int prime[N]; int pn=0; bool vis[N]; LL pow(LL a,LL n,LL mod) { LL base=a,ret=1; while(n) { if(n&1) ret=(ret*base)%mod; base=(base*base)%mod; n>>=1; } return ret%mod; } bool judge(int n) { for(int i=0;prime[i]*prime[i]<=n;i++) { if(n%prime[i]==0) return 1; } return 0; } int main() { for (int i = 2; i < N; i++) { if (vis[i]) continue; prime[pn++] = i; for (int j = i; j < N; j += i) vis[j] = 1; } int a,p; while(~scanf("%d%d",&p,&a),a&&p) { if(!judge(p)){ puts("no"); continue; } if(pow(a,p,p)%p==a) puts("yes"); else puts("no"); } }