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>
#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,many=0,rest;
    scanf("%lld%lld",&n,&k);

    if(n/k>100){
        many=n/1000;
        rest = n;
        n=1000;
    }

    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);
            }
        }
        if(k!=2 && judge==1) a-=k-1;
    }

    if(many!=0){
        ans*=pow(many,3);
        for(a=rest;a>=rest-rest%1000;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);
                }
            }
            if(k!=2 && judge==1) a-=k-1;
        }

    }


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