Trailing zeroes in a Factorial - c

Question

I am trying to write a code for calculating the number of trailing zeroes in a factorial of a specific number (large numbers). However, for small numbers, i get the correct result, but for large the deviations keeps increasing. What's wrong with my logic
#include <stdio.h>
int main(void) {
int t;
scanf("%d", &t);
while (t > 0) {
int factorten = 0, factorfive = 0, factortwo = 0, remainingfive = 0,
remainingtwo = 0;
unsigned int factors = 0;
unsigned int n;
scanf("%u", &n);
for (unsigned int i = n; i > 0; i--) {
if (i % 10 == 0) {
factorten++;
continue;
} else if (i % 5 == 0) {
factorfive++;
continue;
} else if (i % 2 == 0) {
// int new = i;
// while(new % 2 == 0)
//{
// new = new / 2;
factortwo++;
//}
continue;
}
}
factors = factors + factorten;
printf("%u\n", factors);
if (factorfive % 2 == 0 && factorfive != 0) {
factors = factors + (factorfive / 2);
} else {
remainingfive = factorfive % 2;
factors = factors + ((factorfive - remainingfive) / 2);
}
printf("%u\n", factors);
if (factortwo % 5 == 0 && factortwo != 0) {
factors = factors + (factortwo / 5);
} else {
remainingtwo = factortwo % 5;
factors = factors + ((factortwo - remainingtwo) / 5);
}
printf("%u\n", factors);
if ((remainingfive * remainingtwo % 10) == 0 &&
(remainingfive * remainingtwo % 10) != 0) {
factors++;
}
printf("%u\n", factors);
t--;
}
}
Sample Input:
6
3
60
100
1024
23456
8735373
Sample Output:
0
14
24
253
5861
2183837
My OUTPUT
0
13
23
235
5394
2009134

Answer 1

Edit: ignore the first two, they are suboptimal. The third algorithm is optimal.
I think this does what you're trying to do, but is a lot simpler and works:
int tzif(int n)
{
int f2 = 0, f5 = 0;
for (;n > 1; n--)
{
int x = n;
for (;x % 2 == 0; x /= 2)
f2++;
for (;x % 5 == 0; x /= 5)
f5++;
}
return f2 > f5 ? f5 : f2;
}
It counts 2-factors and 5-factors of numbers N...2. Then it returns the smaller of the two (because adding 2-factors is useless without adding 5-factors and vice-versa). Your code is too strange for me to analyze.
I think this should work too, because a factorial will have enough 2-factors to "cover" the 5-factors:
int tzif(int n)
{
int f5 = 0;
for (;n > 1; n--)
for (x = n;x % 5 == 0; x /= 5)
f5++;
return f5;
}
This only counts 5-factors and returns that.
Another method I think should work:
int tzif(int n)
{
int f5 = 0;
for (int d = 5; d <= n; d *= 5)
f5 += n / d;
return f5;
}
Count every fifth number (each has a 5-factor), then every 25-th number (each has another 5-factor), etc.

Answer 2

Have 3 counters - c2,c5,c10.
I think the checks should be
divisible by 5 but not by 10 -> c5++
divisible by 2 but not by 10 -> c2++
divisible by 10. Here if true, then count number of 0's. (c10++)
At last number of 0's will be
smaller_of(c2,c5) + c10
Try to code using this. Should work.

Answer 3

First the trailing 0 in N! are determined by factors 2 and 5 (10). The factors 2 always would be more that the factors 5 in this case you only need to calculate how factors 5 are in the N!.
(N!/5) would give you the number of multiple of 5 (5^1) in N!
(N!/25) would give you the number of multiple of 25 (5^2) in N!
(N!/125) would give you the number of multiple of 125 (5^3) in N!
...
(N!/5^n) would give you the number of multiple of 5^n in N!
When you add the multiple of 5 you are adding too the multiple of 25, 125, ..., 5^n, when you add multiple of 25 you are adding too the multiple of 125, ..., 5^n, etc...
In that case you only need to iterate the power of 5 less or equal than N and add the number of multiple of that 5 power.
Code:
long long trailing_zeros(long long N) {
long long zeros = 0;
for (long long power5 = 5; power5 <= N; power5 *= 5)
zeros += N / power5;
return zeros;
}

Answer 4

#include<iostream>
int main()
{
int size,i;
std::cin >> size;
int*fact;
fact = new int[size];
for (i = 0; i < size; i++)
{
std::cin >> fact[size];
}
for (i = 0; i < size; i++)
{
int con = 5;
int multiple = 0;
do
{
multiple = multiple+(fact[size] / con);
con = con * 5;
} while (con < fact[size]);
std::cout << multiple <<'\n';
}
return 0;
}
this code works perfectly for a single input..bt for multiple inputs it prints the o/p for the last entered number...what is wrong..i jst cant think off it