HDOJ 1905 Pseudoprime numbers（模运算）

 

模运算。。http://www.cnblogs.com/jojoke/articles/1003594.html

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 876    Accepted Submission(s): 358

Problem 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 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 ≤ 1,000,000,000 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

#include<stdio.h>
#include<string.h>
#include<math.h>
__int64 sushu(__int64 a)
{
    __int64 i;
    if(a%2 == 0)
    return 0;
    for(i = 3; i < sqrt(a); i+=2)
    {
        if(a%i == 0)
        return 0;
    }
    return 1;
}
__int64 ab(__int64 a, __int64 p, __int64 x)//模运算公式
{
    __int64 s;
    if(p==0)
    return 1;
    if(p==1)
    return a%x;
    if(p == 2)
    return (a*a)%x;
    s=ab(a,p/2,x);
    if(p%2 == 0)
    return (s*s)%x;
    else
    return (((s*s)%x)*a)%x;

}
int main()
{
    __int64 p, a, x;
    while(scanf("%I64d %I64d",&p,&a) != EOF)
    {
        if(p==0&&a==0)break;
        if(sushu(p)==1)
        printf("no\n");
        else
        {
            x = ab(a,p,p);
            if(x==a)
            printf("yes\n");
            else
            printf("no\n");
        }
    }
    return 0;
}