cses

CSES Solutions (0x6A73)
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commit d5c06c689e3328546cd4c486b30a7909651501ed
parent 19bb82f44e8da0ab854f93248a8fe72d430b46e4
Author: Jack Sangdahl <jack@pngu.org>
Date:   Thu, 13 Feb 2025 15:18:28 -0700

1618. trailing zeors

Diffstat:
Mintroductory_problems/readme.md | 7+++++++
Aintroductory_problems/trailing_zeros.cpp | 13+++++++++++++
2 files changed, 20 insertions(+), 0 deletions(-)

diff --git a/introductory_problems/readme.md b/introductory_problems/readme.md @@ -62,3 +62,10 @@ Your task is to calculate the number of bit strings of length `n`. For example, **Input**: The only input line has an integer `n`. (Example: 3) **Output**: Print the result modulo 10E9 + 7. (Example: 8) + +### Trailing Zeros +Your task is to calculate the number of trailing zeros in the factorial `n!`. For example, `20! = 20!=2432902008176640000` and it has 4 trailing zeros. + +**Input**: The only input line has an integer `n`. (Example: 20) + +**Output**: Print the number of in `n!`. (Example: 4) diff --git a/introductory_problems/trailing_zeros.cpp b/introductory_problems/trailing_zeros.cpp @@ -0,0 +1,13 @@ +// 1618. Trailing Zeros +#include <bits/stdc++.h> +using namespace std; + +int main() { + static int n, ans; + scanf("%d", &n); + while (n) { + n /= 5; + ans += n; + } + printf("%d\n", ans); +}