题目←

头一次打那么长的代码re了一遍就过了！（哭）

唔……略微的眼熟，跟清北学堂那天的模拟有点像？
总之找规律吧……找找找找……
等等为什么代码这么长？
是不是处理了很多没用的东西？算了不管了……

#include<iostream>
#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std;
const LL MAXN = 100000 + 50;
struct zt
{
    LL l,r;
}lim[MAXN];
struct zt2
{
    LL u,d;
}h[MAXN];
LL n,L[MAXN],R[MAXN];
LL U[MAXN],D[MAXN];
LL find(LL x,LL y)
{
    if(x <= n/2)
    {
        if(y >= lim[x].l && y <= lim[x].r)
        {
            return L[x] + (y - lim[x].l);
        }
        else if(y < lim[x].l)
        {
            return U[y] - (x - h[y].u);
        }
        else if(y > lim[x].r)
        {
            return U[y] + (x - h[y].u);
        }
    }
    else
    {
        if(y >= lim[x].l && y <= lim[x].r)
        {
            return L[x] - (y - lim[x].l);
        }
        else if(y < lim[x].l)
        {
            return U[y] + (h[y].u - x);
        }
        else if(y > lim[x].r)
        {
            return U[y] - (h[y].u - x);
        }
    }
}
LL k;
int main()
{
    scanf("%lld%lld",&k,&n);
    LL tmp = 0;
    for(LL i = 1;i <= n/2;i ++)
    {
        lim[i].l = 1 + tmp;
        lim[i].r = n - tmp;
        tmp ++;
    }
    if(n % 2)
    {
        lim[n/2 + 1].l = n/2 + 1;
        lim[n/2 + 1].r = n/2 + 1;
    }
    tmp = 0;
    for(LL i = n;i > n - n/2;i --)
    {
        lim[i].l = 1 + tmp;
        lim[i].r = n - tmp;
        tmp ++;
    }
    tmp = n - 1;
    LL pre1 = 1,pre2 = n;
    L[1] = pre1;R[1] = pre2;
    LL temp = 1;
    U[1] = L[1] + 3 * (n - temp) + n - temp - 1;
    U[n] = R[1];
    temp += 2;
    for(LL i = 2;i <= n/2;i ++)
    {
        L[i] = pre1 + 4 * tmp;
        R[i] = pre2 + 4 * tmp - 2;
        U[i] = L[i] + 3 * (n - temp) + n - temp - 1;
        U[n - i + 1] = R[i];
        pre1 = L[i];
        pre2 = R[i];
        tmp -= 2;
        temp += 2;
    }
    if(n%2)
    {
        L[n/2 + 1] = n*n;
        R[n/2 + 1] = n*n;
        U[n/2 + 1] = n*n;
        D[n/2 + 1] = n*n;
    }
    pre1 = 1 + 3 * (n - 1);pre2 = 1 + 2 * (n - 1);
    LL tmp1 = 2;
    LL tmp2 = 3;
    L[n] = pre1;R[n] = pre2;
    D[n] = L[n];
    for(LL i = n - 1;i > n - n/2;i --)
    {
        L[i] = pre1 + (n - tmp1) * 2 + (n - tmp2) * 2;
        R[i] = L[i] - (n - (n - i) * 2) + 1;
        D[n - i + 1] = L[i];
        D[i] = R[i];
        pre1 = L[i];
        pre2 = R[i];
        tmp1 += 2;
        tmp2 += 2;
    }
    tmp = 0;
    for(LL i = 1;i <= n/2;i ++)
    {
        h[i].u = 1 + tmp + 1;
        h[i].d = n - tmp;
        tmp ++;
    }
    if(n % 2)
    {
        h[n/2 + 1].u = n/2 + 1;
        h[n/2 + 1].d = n/2 + 1;
    }
    tmp = 0;
    for(LL i = n;i > n - n/2;i --)
    {
        h[i].u = 1 + tmp;
        h[i].d = n - tmp;
        tmp ++;
    }
    LL a,b;
    for(LL i = 1;i <= k;i ++)
    {
        scanf("%lld%lld",&a,&b);
        printf("%lld\n",find(a,b));
    }
    return 0;
}