在使用大数 C++ 时检查返回返回 0 的最大质因数的函数

Question

This is for the third problem in Project Euler. I suspect that it only returns 0 because I assigned 0 to the 'returnum' variable on initialization. My code works fine in smaller numbers, but it isn't working for a large number (600851475143) which is needed to get the correct answer.

Is this for the size of int? If so what data type should I use?

Here is my code:

int problem3(long int num) {
    long int returnum;
    for (int i = 2; i < num; ++i) {
        if (num % i == 0 && primecheck(i)) {
           returnum = i;
        }
    }
    return returnum;
}

And this is my 'primecheck' function:

bool primecheck(long int num) {
     for (int i = 2; i <= num / 2; ++i) {
         if (num % i == 0) {
            return false;
        }
    }
    return true;
}

Edit: I have tried to change the datatype but when I do, it just doesn't respond, showing an empty console for a few minutes, then crashing.

Answer 1

The maximum value for a variable of type int is 2147483647, whereas the max value of float is : 3.40282e+38 or 0x1.fffffep+127. You could use double also.

Answer 2

long long problem3(long long num) {
long long returnum;
for (long long i = 2; i < num; ++i) {
    if (num % i == 0 && primecheck(i)) {
       returnum = i;
    }
}
return returnum;
}

And this is my 'primecheck' function:

bool primecheck(long long num) {
     for (long long i = 2; i <= num / 2; ++i) {
         if (num % i == 0) {
            return false;
        }
    }
    return true;
}

Fundamental Types (C++)

tested! OK! VS2017

#include "stdafx.h"
#include <stdio.h>

bool primecheck(long long num);
long long problem3(long long num);

bool primecheck(long long num) {
    for (long long i = 2; i <= num / 2; ++i) {
        if (num % i == 0) {
            return false;
        }
    }
    return true;
}

long long problem3(long long num) {
    long long returnum;
    for (long long i = 2; i < num; ++i) {
        if (num % i == 0 && primecheck(i)) {
            returnum = i;
        }
        if (i % 100000 == 0) {
            printf("tick - %lld\n", i);
        }
    }
    return returnum;
}


int main()
{
    printf("out number %lld\n", problem3(600851475143));
    return 0;
}

32 037 210 000 - screenshot 10 000 000 per/sec . (in this moment > (long)2^32) and what you want with this program? wait 65K sec? (18 hours) may be need use math algorithm for best time ?