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)

//わかんね
#include<stdio.h>
typedef long long int ll;

ll conbi(ll a, ll b){
    if(a-b == 0){return 1;}
    ll p1 = 1;
    ll sa = a;
    for(int i = 0; i < b; i++){
        p1 *= sa;
        sa--;
    }
    //printf("p1:%lld\n", p1);
    
    return p1/b;
}

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

    //N以下で、かつKの倍数の数字の個数をmultiofkに入れる
    ll multiOfk = 0;
    for(ll i = 1; i <= N; i++){
        if(i%K == 0){multiOfk++;}
    }
    
    ll ans = multiOfk;
    //printf("ans1:%d\n", ans);
    if(multiOfk >= 2){ans += conbi(multiOfk,2)*3*2;}
    //printf("ans2:%d\n", ans);
    //printf("2帰ってきた\n");
    if(multiOfk >= 3){ans += conbi(multiOfk,3)*3*2;}
    //printf("ans3:%d\n", ans);
    //printf("3帰ってきた\n");
    //return 0;
    //cntは、NをKで割ったあまりがK/2となる数字の個数をいれる
    ll cnt = 0;
    if(K%2 == 0){
        ll KK = K/2;
        while(N > (cnt*K + KK)){cnt++;}
    }

    ans += cnt;
    //printf("ans+=cnt:%lld\n", ans);
    //return 0;
    if(cnt >= 2){ans += conbi(cnt,2)*3*2;}
    if(cnt >= 3){ans += conbi(cnt,3)*3*2;}

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

}