And demand factors
Subject description:
Factors and do not include all of its own and a number of factors, such as Factor 12 are so 1,2,3,4,6 factor 12 and 16. Now given a number n (n <= 10 ^ 9), and find its factor.
Input formats:
A number.
Output formats:
A number.
Sample input:
12
Sample output:
16
prompt:
60% of the data, n-<10,000 =;
80% of the data, n-<= 1, 000,000;
100% data, n <= 1,000,000,000;
This problem first written when most people think of direct enumeration, enumeration that is, from 1 up to the maximum number of direct factor in addition to itself. So with this piece of code below:
#include <stdio.h>
int main()
{
long n, s = 0, x;
int i;
scanf("%ld", &n);
for (i = 1; i < n; i++) {
if (n % i == 0)
s += i;
}
printf("%ld", s);
return 0;
}
But this code is encountered in large numbers on a timeout. The reason is that from 1 up to too many to enumerate the maximum number of factors, efficiency is too low. There is a solution, that is, when the number of enumeration a factor, flew input is divided by the factor, so that you can only enumerate to the square root of n. code show as below:
#include <stdio.h>
#include <math.h>
int main()
{
long n, s = 0;
int i;
scanf("%ld", &n);
for (i = 1; i <= sqrt(n); i++) {
if (n % i == 0)
if (i == n / i)
s += i;
else
s = s + i + n / i;
}
printf("%ld", s - n);
return 0;
}
