[C++知识库]Codeforces Round #768 (Div. 2) C. And Matching

思路：
如果k不为n-1的话，就把k和n-1配，0和n-1-k配，其他相加为n-1就配一对
如果为n-1的话就把后四组和前四组单独配一下，其他相加为n-1配一对,尽量不要打乱中间的配对
Code

#include <bits/stdc++.h>
// #define DEBUG freopen("_in.txt", "r", stdin);
#define DEBUG freopen("_in.txt", "r", stdin), freopen("_out.txt", "w", stdout);
typedef long long ll;
using namespace std;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 1e6 + 10;
const ll maxm = 1e7 + 10;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;

ll T, n, k;
ll arr[maxn], brr[maxn];

int main()
{
    // DEBUG;
    // printf("%lf",pow(2,16));
    scanf("%lld", &T);
    while (T--)
    {
        scanf("%lld%lld", &n, &k);
        if (k == n - 1)
        {
            if (n == 4)
                printf("-1\n");
            else
            {
                printf("%lld %lld\n", n-2, n - 1);
                printf("0 2\n");
                printf("1 %lld\n", n-3);
                printf("3 %lld\n", n-4);
                for (ll i = 4; i < n / 2; i++)
                {
                    printf("%lld %lld\n", i, n - 1 - i);
                }
            }
            continue;
        }
        printf("%lld %lld\n", k, n - 1);
        if (k != 0)
            printf("0 %lld\n", n - 1 - k);
        for (ll i = 0; i < n / 2; i++)
        {
            if (i != k && i != 0 && i != n - 1 - k && i != n - 1 && n - 1 - i != k && n - 1 - i != 0 && n - 1 - i != n - 1 - k && n - 1 - i != n - 1)
            {
                printf("%lld %lld\n", i, n - 1 - i);
            }
        }
        // printf("\n");
    }

    return 0;
}