SOJ 1009. Mersenne Composite N

题目大意：梅森数定义为2^k - 1 的数，梅森合数定义为由素数相乘获得的梅森数。输入n( n ≤ 63), 找出少于n的素数k，使得2^k - 1为梅森合数，并且找到该梅森合数的素数因子。

解题思路：可以先写一个程序找出素数k，使得梅森数2^k - 1不是素数。这些k分别是11,23,29,37,41,43,47,53,59。然后找到对应的素数因子即可。也可以不预先找到目标素数k.

代码如下：

#include <iostream>
#include <vector>
#include <ctime>
using namespace std;

const int maxn = 64;
vector<int> primes;
bool arr[maxn];

int n;

void init() {
    for (int i = 0; i < maxn; i++) {
        arr[i] = true;
    }

    for (int i = 2; i < maxn; i++) {
        if (arr[i]) {
            primes.push_back(i);
            for (int j = 2; i * j < maxn; j++) {
                arr[i * j] = false;
            }
        }
    }
}

int main() {
    cin >> n;
    double start = clock();
    init();
    for (int i = 0; primes[i] <= n && primes[i] <= 59; i++) {
        vector<long long> vt;

        long long m = ((long long)1 << primes[i]) - 1;

        for (long long i = 2; i * i <= m; i++) {
            if (m % i == 0) {
                vt.push_back(i);
                m = m / i;
            }
        }

        if (m != 1) {
            vt.push_back(m);
        }

        if (vt.size() > 1) {
            cout << vt[0];
            for (int i = 1; i < vt.size(); i++) {
                cout << " * " << vt[i];
            }
            cout << " = " << ((long long)1 << primes[i]) - 1 << " = ( 2 ^ " << primes[i] << " ) - 1" << endl;
        }

    }
    double end = clock();
    cout << (end - start) / CLOCKS_PER_SEC << endl;

    return 0;
}