Pseudoprime numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11470 Accepted: 4947
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 ≤ 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
解题心得：
- 题意就是给你两个数a,b，如果a的b次方 mod p等于a，并且p不是素数那么输出yes，否则输出no
- 先素数筛选判断p是不是素数，然后一个快速幂模拟一下最后mod是否等于a就行了，其实就是一个费马小定理的逆命题否定判断，就这个题来说学没学过数论都很简单。
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll mod_mult(ll x,ll y) {
ll res = 1;
ll p = y;
while(y) {
if(y&1)
res = (res*x)%p;
x = (x*x)%p;
y >>= 1;
}
return res%p;
}
bool checke_prim(int n) {
for(int i=2;i*i<=n;i++)
if(n % i == 0)
return true;
return false;
}
int main() {
ll p,a;
while(scanf("%lld%lld",&p,&a) && p|a) {
bool flag = checke_prim(p);
ll ans = mod_mult(a,p);
if(ans == a && flag)
printf("yes
");
else
printf("no
");
}
return 0;
}