doruk - 2 months ago 8x

C Question

Problem: Find the number of integers 1 < n < 10^7, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.

I can't reach 10^7 because it is too big number for C and me. How can i solve this problem in C?

`#include<stdio.h>`

#include<conio.h>

int divisorcount(int);

int main()

{

int number,divisornumber1,divisornumber2,j=0;

for(number=1;number<=100;number++){

divisornumber1=divisorcount(number);

divisornumber2=divisorcount(number-1);

if(divisornumber1==divisornumber2){

printf("%d and %d\n",number-1,number);

j++;

}

}

printf("\nThere is %d integers.",j);

getch();

}

int divisorcount(int num)

{

int i,divi=0;

for(i=1;i<=(num)/2;i++)

if(num%i==0)

divi++;

return divi;

}

Answer

Ever tried `long long num = 100000000LL;`

? C isn't smart enough to conclude the type on the right side from the left `long long`

so you have to add the `LL`

. With this approach you should be able to handle larger numbers than normal integers, just change your functions and variables in a suitable way.

A `long long`

is always at least `2^64`

bit in size which you can check on Wikipedia.

*Hint:* As someone mentioned in the comments, Project Euler is not about bruteforcing. This is a lame approach. Think about some better strategies. You might want to get help at math.stackexchange?

**EDIT:** I don't know why I thought, that a `uint32_t`

is not enough for `10^7`

- sorry for that mistake.

Source (Stackoverflow)

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