Problem:
Score : 300 points

Problem Statement

You are given integers N and K. Find the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.
The order of a,b,c does matter, and some of them can be the same.

Constraints

1 \leq N,K \leq 2\times 10^5

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.

Sample Input 1

3 2

Sample Output 1

9

(1,1,1),(1,1,3),(1,3,1),(1,3,3),(2,2,2),(3,1,1),(3,1,3),(3,3,1) and (3,3,3) satisfy the condition.

Sample Input 2

5 3

Sample Output 2

1

Sample Input 3

31415 9265

Sample Output 3

27

Sample Input 4

35897 932

Sample Output 4

114191

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::HashSet;
use std::cmp;
use std::f64::consts;

fn main() {
    let n: i32  = read();
    let k: i32 = read();
    let mut count = 0;

    if k % 2 == 0 {

        let mut count1: i32 = 0;
        // kで割ったあまりが0の場合
        for i in 1..n+1 {
            if i % k == 0 {
                count1+=1;
            }
        }
        count+=count1.pow(3);

        let mut count2: i32 = 0;
        // 全部 kで割ったあまりが k/2 の場合
        for i in 1..n+1 {
            if i % k == k/2 {
                count2+=1;
            }
        }

        // 全部 k / 2の場合
        count += count2.pow(3);
    } else {
        let mut count1: i32 = 0;
        // kで割ったあまりが0の場合
        for i in 1..n+1 {
            if i % k == 0 {
                count1+=1;
            }
        }
        count+=count1.pow(3);
    }

    println!("{}", count);
}

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

// 最大公約数
fn gcd(a: i32, b: i32) -> i32 {
    match b {
        0 => a,
        _ => gcd(b, a % b)
    }
}

// 最小公倍数
fn lcm(a: i32, b: i32) -> i32 {
    a * b / gcd(a, b)
}

// 階乗
fn kaijou(n: i32)->i32 {
    if n == 1 {
        return n;
    }
    return n * kaijou(n-1);
}