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: Wrong Answer)

#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);

    //それぞれN以下の自然数の中で、もともとKの倍数と2倍して初めてKの倍数になるものの数
    int k1=0;
    int k2=0;
    for(int i=1;i<=N;i++){
        if(i%K==0) k1++;
        else if((i*2)%K==0) k2++;
    }


    long ans=(k1*k1*k1)+(k2*k2*k2);
    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