首页 > 代码库 > poj 3641 Pseudoprime numbers 【快速幂】
poj 3641 Pseudoprime numbers 【快速幂】
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6645 | Accepted: 2697 |
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：给出两个数p，a 如果p是素数输出no 否则判断a的p次方对p取模之后是不是等于a
代码：
#include <stdio.h>
#include <math.h>
#define LL __int64
int is_prime(int n){
if(n < 2) return 0;
for(int i = 2; i <= sqrt(n+0.0); i++){
if(n%i == 0) return 0;
}
return 1;
}
int fast(int p, int a){
LL r = p, t = 1, mod = p;
while(r > 0){
if(r&1) t = ((t%mod)*(a%mod))%mod;
a = ((a%mod)*(a%mod))%mod;
r >>= 1;
}
return t%mod;
}
int main(){
int p, a;
while(scanf("%d%d", &p, &a), p||a){
if(is_prime(p)){
printf("no\n"); continue;
}
else if(fast(p, a) == a){
printf("yes\n");
}
else printf("no\n");
}
return 0;
} poj 3641 Pseudoprime numbers 【快速幂】