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)

#pragma GCC diagnostic ignored "-Wunused-result"

#include <stdio.h>

int main(void) {
  int N,K,a,b,c;
  int count = 0;
  static int pair[200000] = {0};
  
  scanf("%d %d", &N, &K);

  int max = (N>K)?K:N;
  // printf("max %d\n",max);
  a = K;
  b = K;
  c = K;

  int mul;
  count = 1;
  if (K%2==0) {
    mul=N/K;
    count += 1;
    count += mul*3 + mul*mul*3 + mul*mul*mul;
  }
  mul=(N-K)/K;
  if (mul>1) {
    count += mul*3 + mul*mul*3 + mul*mul*mul;
  }
  /* for (int i=1;i<=max;i++) { */
  /*   for (int j=1;j<=max;j++) { */
  /*     if ((i+j)%K == 0) pair[i]=j; */
  /*   } */
  /* } */
  /* for (int i=1; i<=max; i++) { */
  /*   int a=i; */
  /*   int b=pair[a]; */
  /*   int c=pair[b]; */
  /*   if ((c+a)%K==0) { */
  /*     // printf("%d,%d,%d\n",a,b,c); */
  /*     int mul=(N-a)/K; */
  /*     count += 1 + mul*3 + mul*mul*3 + mul*mul*mul; */
  /*   } */
  /* } */
  
  /* int a,b,c; */
  /* a=1; b=1; c=1; */
  /* for (a=1;a<=N;a++) { */
  /*   if ( (a+b)<K || */
  /*        (c+a)<K) continue; */
  /*   for (b=1;b<=N;b++) { */
  /*     if ( (a+b)<K || */
  /*          (b+c)<K) continue; */
  /*     for (c=1;c<=N;c++) { */
  /*       if ( (b+c)<K || */
  /*            (c+a)<K) continue; */
  /*       if ( ((a+b)%K==0) && */
  /*            ((b+c)%K==0) && */
  /*            ((c+a)%K==0) ) { */
  /*         //printf("%d,%d,%d\n",a,b,c); */
  /*         count++; */
  /*       } */
  /*     } */
  /*   } */
  /* } */

  printf("%d\n",count);
  return 0;

}