pojPseudoprime numbers (快速幂)

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 p^th 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

no
no
yes
no
yes
yes

这道题的意思是输入两个数p,a，当满足p不是素数且pow(a,p)%p==a时输出yes,否则输出no
思路：只要会最基本的快速幂就可以做出这道题来。
快速幂：思路就是将pow(a,p)中的p换成多个2的倍数相加的形式，那么pow(a,p)=pow(a,c1)*pow(a,c2)*....*pow(a,cn),其中p=c1+c2+..+cn;利用位运算符
可以很容易实现这种操作。

int Fastpow(int a,int p){

int Fastpow(int a,int p){
long long int base=a;
long long int res=1;
while(p){
if(p&1) //若p的二进制表示形式最后一位数是1，则为真，否则为假
res*=base/*,res=res%mod*/;
base*=base;
/*base=base%mod;*/
p>>1; //将p的二进制表示形式后移一位，把刚处理过的p的最后一位去掉
}
return res;
}

快速幂的第二种表达方式就是利用递归

int Fastpow(int a,int p){
if(p==1) return a;
long long int temp=Fastpow(a,p/2)%p;
if(p%2==1) return temp*temp%p*a%p; //这里的第一个p若是省略，则会wrong answer
else return temp*temp%p;
}

最后附上AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
long long int p;
int Fastpow(int a,int n){
long long int m=n,base=a;
long long int res=1;
while(m){
if(m&1){
res=res*base%p;
}
base=(base*base)%p;
m>>=1;
}
return res;
}
/*int Fastpow(int a,int n){
if(n==1) return a;
long long int temp=Fastpow(a,n/2)%p;
if(n%2==1) return temp*temp%p*a%p;
else return temp*temp%p;
}*/
int main(){
long long int a,c[100000];
while(~scanf("%lld%lld",&p,&a)){
if(p==0&&a==0) return 0;
int plug=0;
for(int i=2;i*i<=p;i++)
if(p%i==0) plug=1;
if(plug==0) {
printf("no\n");
continue;
}
//cout<<Fastpow(a,p)<<endl;
if(Fastpow(a,p)==a) printf("yes\n");
else printf("no\n");
}
}

这个是最基本的判断是否是素数，利用的是试除法，下面给出一个利用素数筛的快捷算法，比上面这一个耗时少不少：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=31700;
const long long int maxn1=1e9+10;
int Fastpow(int a,int p){
long long int m=p,base=a;
long long int res=1;
while(m){
if(m&1){
res=res*base%p;
}
base=(base*base)%p;
m>>=1;
}
return res;
}
int main(){
long long int p,a;
int c[maxn],prim[maxn],j=0;
memset(c,0,sizeof(c));
for(int i=2;i*i<maxn1;i++){
if(c[i]==0){
prim[j]=i,j++;
for(int k=i*i;k<maxn;k+=i)
c[k]=-1;
}
}
while(~scanf("%lld%lld",&p,&a)){
if(p==0&&a==0) return 0;
int plug=1;
for(int i=0;i<j&&prim[i]<p;i++)
if(p%prim[i]==0) plug=0;
if(plug==1) {
printf("no\n");
continue;
}
if(Fastpow(a,p)==a) printf("yes\n");
else printf("no\n");
}
}

pojPseudoprime numbers (快速幂)

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