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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

//昇順
int asc(const void* a, const void* b){
    return *(int*)a-*(int*)b;
}

//降順
int desc(const void* a, const void* b){
    return *(int*)b-*(int*)a;
}

int main(void){
    int N,K;
    scanf("%d%d",&N,&K);

    //N以下の自然数の中で、2倍したらKの倍数になるものの数
    long ans=0;
    for(int a=1;a<=N;a++){
        if((a*2)%K==0){
            for(int b=1;b<=N;b++){
                if((b*2)%K==0 && (a+b)%K==0){
                    for(int c=1;c<=N;c++){
                        if((a+c)%K==0 && (b+c)%K==0) ans++;
                    }
                }
            }
        }
    }

    printf("%ld\n",ans);

    return 0;
}

//a+b,b+c,c+aがKの倍数→2*a,2*b,2*cがKの倍数