Programming Algorithm-minimum number of code divisible by 1 to n (c)

Code that can be divisible by 1 to n at least (c)

Address: http://blog.csdn.net/caroline_wendy

The minimum integer that can be divisible by 1 to n isProduct of all prime numbers from 1 to n.

Evaluate all prime numbers from 1 to n. The prime number factor with the maximum number can only beSQRT (N)To reduce the traversal range.

The time complexity is O (n). O (SQRT (n) * SQRT (n )).

Code:

/* * main.cpp * *  Created on: 2014.7.20 *      Author: Spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <memory.h>#include <math.h>#include <iostream>#include <vector>using namespace std;void Primes (int n, vector<int>& vi) {if (n <= 2)return;bool* a = new bool[n+1];for (int i=1; i<=n; i++)a[i] = true;for (int i=2; i<sqrt(n); i++) {for (int j=2; j*i<=n; j++) {a[i*j] = false;}}for (int i=1; i<=n; ++i)if (a[i])vi.push_back(i);delete[] a;}int main(void){vector<int> vi;int n = 3;Primes(n, vi);long long min = 1;for (size_t i=0; i<vi.size(); ++i) {min *= vi[i];}cout << "min = " << min << endl;return 0;}

Output:

min = 30030

Programming Algorithm-minimum number of code divisible by 1 to n (c)