B - B
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Submit
Status
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:给定数a,p使a^p%p=a,其中p不是素数。此题介绍的就是关于伪素数的知识,题中的公式即费马小定理。
- 思路:用快速幂求余数,用素数判定法判断是否为素数。
- 失误:英文题,理解有点模糊,刚开始没注意p不能是素数。
- 代码如下:
#include<cstdio>
#include<cmath>
__int64 quickpow(__int64 base,__int64 m,__int64 mod)//快速幂取模
{
__int64 ans=1;
while(m>0)
{
if(m%2==1)
{
ans=(ans*base)%mod;
--m;
}
base=(base*base)%mod;
m/=2;
}
return ans;
}
__int64 prime(__int64 n)//是否是素数
{
__int64 i;
if(n%2==0)//一步就可以筛除一半的数,用数论的知识优化代码,提高速度
return 0;
for(i=2;i<= sqrt(n);++i)
{
if(n%i==0)
return 0;
}
return 1;
}
int main()
{
__int64 a,p,c;
while(scanf("%I64d %I64d",&p,&a)!=EOF&&(a||p))
{
c=quickpow(a,p,p);
if(c==a&& (prime(p)==0))
printf("yes\n");
else
printf("no\n");
}
return 0;
}