[Project Euler] Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

#include <iostream>
using namespace std;

int prime[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,
                43,47,53,59,61,67,71,73,79,83,89,97, 
                101,103,107,109,113,127,131,137,139,
                149,151,157,163,167,173,179,181,191,
                193,197,199,211,223,227,229,233,239,
                241,251,257,263,269,271,277,281,283,
                293,307,311,313,317,331,337,347,349, 
                353,359,367,373,379,383,389,397,401,
                409,419,421,431,433,439,443,449,457,
                461,463,467,479,487,491,499,503,509,
                521,523,541,547,557,563,569,571,577,
                587,593,599,601,607,613,617,619,631,
                641,643,647,653,659,661,673,677,683,
                691,701,709,719,727,733,739,743,751,
                757,761,769,773,787,797,809,811,821,
                823,827,829,839,853,857,859,863,877,
                881,883,887,907,911,919,929,937,941,
                947,953,967,971,977,983,991,997};

bool isPrime(int n);
            
int main(){
    long long num = 600851475143;
    
    for(int i=799999; i>1; i-=2){
        if(num%i==0 && isPrime(i)){
            cout << i << endl;
            break;
        }
    }
    return 0;
}


bool isPrime(int n){
    for(int i=0; i<168; i++){
        if(n%prime[i] == 0)
            return false;
    }
    return true;
}

好啦，得到的结果是正确的

但其实这个算法是错误的

在验证一个数是否是素数时，我们一般是从2到sqrt（）

800000接近于600851475143的平方根

但实际上求解600851475143的最大质因数不应该认为该质因数小于sqrt（600851475143）

如15的最大质因数是5，但5不小于sqrt（15）

所以，改到下面的算法

#include <iostream>
using namespace std;
            
int main(){
    long long num = 600851475143;
    int divisor = 3;
    
    while( num>1){
        if( num%divisor== 0){
            num /= divisor;
            divisor -= 2;
        }
        divisor += 2;
    }
    cout << divisor << endl;
    return 0;
}

该算法思路更清晰

把因数从小到大一一分解出去，最后分解的因数必然是最大的因数

我们也可以把divisor声明为long long型的。

因为divisor可能超过了int的范围