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>
#include<math.h>
#include<string.h>
#include <time.h>
typedef long long llint;

int main(void){
    llint n,i,k,ans=0,a,b,c,judge=0;
    scanf("%lld%lld",&n,&k);
  
    for(a=n;a>=1;a--){
        for(i= (2*a%k==0)? 2*a/k: 2*a/k-1 ;i>0;i--){
            judge=0;
            b = k*i-a;
            if(a<b) break;
            if(b<1) break;
            c=b;
            if((c+b)%k==0){
                llint div;
                if(a==b){
                    if(a==c) ans++;
                    if(b!=c) ans+=3;
                    div = c%k==0?c/k-1:c/k;
                    if(div<1) continue;
                    ans+=3*div;
                }else{
                    if(b!=c) ans+=6;
                    if(b==c ) ans+=3;
                    div = c%k==0?c/k-1:c/k;
                    if(div<1) continue;
                    ans+=6*div;
                }
                judge++;
                // printf("%lld %lld %lld %lld %lld\n",a,b,c,ans,div);
            }
            
            // for(c=b;b+c>0;c-=k){
            //     if(c<1) break;
            //     if((c+b)%k==0){
            //         if(a==b && a==c) ans++;
            //         if((a==b && a!=c) || (b==c && a!=b)) ans+=3;
            //         if(a!=b && b!=c ) ans+=6;
            //         judge++;
            //         // printf("%lld %lld %lld %lld\n",a,b,c,ans);
            //     }
            // }
        }
        if(k!=2 && judge==1) a-=k-1;
    }



    printf("%lld\n",ans);
    return 0;
}