題意：

給你兩個數，p和a；滿足兩個條件：
1.p不是素數；
2.$a^p$ %p==a;
滿足則輸出yes，反之輸出no。

題目：

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 ≤ 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

分析：

這道題一旦題意出來了，其他就是常規(guī)操作，第一個條件p為非素數很容易看出，但第二個條件，恕我眼拙，題目給了一個【a > 1, $a^p$ = a (mod p). 】的條件，我很容易就想到了費馬小定理的$x^{p?1}\equiv 1(modp)$,然后就掉進坑里爬都爬不起來，看到博客上的題意，心情很崩潰。

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; ll p,a; bool prime(ll x){int k=0;for(int i=2;i*i<=x;i++)if(x%i==0)k++;if(k==0) return true;return false; } ll mod(ll a,ll p,ll mod){ll ans=1;a%=mod;while(p){if(p&1) ans=a*ans%mod;p>>=1;a=a*a%mod;}return ans; } int main(){while(~scanf("%lld%lld",&p,&a)){if(!p&&!a)break;if(prime(p)){printf("no\n");continue;}if(mod(a,p,p)==a) printf("yes\n");else printf("no\n");}return 0; }

總結

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