Project Euler 97 ：Large non-Mersenne prime 非梅森大素数

Large non-Mersenne prime

The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 2^6972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p−1, have been found which contain more digits.

However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×2^7830457+1.

Find the last ten digits of this prime number.

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非梅森大素数

1999年人们发现了第一个超过一百万位的素数，这是一个梅森素数，可以表示为2^6972593−1，包含有2,098,960位数字。在此之后，更多形如2^p−1的梅森素数被发现，其位数也越来越多。

然而，在2004年，人们发现了一个巨大的非梅森素数，包含有2,357,207位数字：28433×2^7830457+1。

找出这个素数的最后十位数字。

解题

感觉很简单。。。

JAVA

package Level3;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.math.BigInteger;
import java.util.ArrayList;

public class PE097{
    public static void run() {
        BigInteger m = new BigInteger("10000000000");
        BigInteger r1 = new BigInteger("28433");
        BigInteger t = new BigInteger("2");
        BigInteger exp = new BigInteger("7830457");
        BigInteger res = t.modPow(exp, m);
        res = r1.multiply(res).add(new BigInteger("1"));
        res = res.mod(m);
        System.out.println(res);
    }
    public static void main(String[] args) throws IOException {
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");

    }
}

// 8739992577
// running time=0s2ms

就这样

或者这样

    public static void run2(){
        long base = 2;
        long mod = 1000000000;
        long exp = 7830457;
        long res = 28433;
        for(long i =1;i<=exp;i++){
            res = (res*2)%mod;
        }
        res +=1;
        res %=mod;
        System.out.println(res);
    }
//    739992577
//    running time=0s163ms

上面mod少个0求的是后9位的数，因为多个0就越界了，少一位手工0到9可以暴力遍历。。。

Python

# coding=gbk
import copy
import time as time
def main():
    print ((28433*(2**7830457))+1)%10000000000
t0 = time.time()
main()
t1 = time.time()
print "running time=",(t1-t0),"s"
#
# running time= 0.0190000534058 s

也就这样