NEFU120 Mason Prime number "Lucas-lehmer judgment"

Topic Links:

http://acm.nefu.edu.cn/JudgeOnline/problemshow.php?problem_id=120

Main topic:

Give a number p, if MP = 2^p-1 is a mason prime, then output "yes", otherwise output "no".

Ideas:

This problem p is less than 62. MP is less than 2^62-1. If the direct number of primes will be relatively slow, using the Sieve method of the amount of data

It's going to be big. So here is the rule of lucas-lehmer judgment.

Lucas-lehmer decision rule Specific steps:

If you want to determine the integer MP = 2^p-1, make lucas sequence date[1] = 4,l (i) = (L (i-1) ^2-2)% MP,

If date[p-1] = = 0, then the number of MP is Mason Prime. In particular, the first Mason prime number 2 is to be specially sentenced.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    Std;long Long Date[70];long long Multi (Long long A,long long B,long long m)//a * B% m{long long ret = 0;        while (b) {if (b & 1) ret = (ret + a)% m;        b >>= 1;    A = (a << 1)% m; } return ret;}    int main () {Long long sum,temp;    int t,n;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);        sum = 1;        sum = (sum<<n);        Sum-= 1;        DATE[1] = 4;            for (int i = 2; I <= N-1; ++i) {temp = Multi (date[i-1],date[i-1],sum);        Date[i] = (temp-2)% sum;        } if (N = = 2) printf ("yes\n");            else {if (date[n-1] = = 0) printf ("yes\n");        else printf ("no\n"); }} return 0;}

NEFU120 Mason Prime number "Lucas-lehmer judgment"