牛客网多校第七场 E Counting 4-Cliques

/*
题意：输入k(<=1e6)，构造一个图使得图中大小为4的团恰好有k个。

首先考虑构造完全图，那么t个点的完全图一共能有C(t, 4)个大小为4的团。但是在C(t, 4)和C(t+1, 4)之间会有空缺，因此在完全图外放若干个点，每个点与这个完全图中的若干个点连边，最后会形成类似于C(t, 4)+C(x1, 3)+C(x2, 3)+…这样的式子，并且要能补满中间的空隙。

至于为什么是另外找五个点，没办法证明出来，但四个点的话 额外的17个大小为4的团就出不来
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;

ll C4(ll n) { // C(n, 4)
    ll ans = n * (n - 1) * (n - 2) * (n - 3) / (1 * 2 * 3 * 4);
    return ans;
}

ll C3(ll n) { // C(n, 3)
    ll ans = n * (n - 1) * (n - 2) / (1 * 2 * 3);
    return ans;
}

ll mp[200000 + 5];
int k;

int main() {
    memset(mp, -1, sizeof(mp));
    for (ll i = 1; i <= 100; i++) {
        mp[C3(i)] = i;
    }
    scanf("%d", &k);
    ll C = 70;
    ll t = 4;
    while ((t + 1) <= C && C4(t + 1) <= k) {
        t++;
    }
    C = min(C, t);
    ll a, b, c, d, e = -1;
    for (a = 2; a <= C; a++) {
        for (b = a; b <= C; b++) {
            for (c = b; c <= C; c++) {
                for (d = c; d <= C; d++) {
                    ll cnt = C3(a) + C3(b) + C3(c) + C3(d);
                    if (cnt <= k - C4(t)
                            && mp[k - C4(t) - cnt] >= 0
                            && mp[k - C4(t) - cnt] <= C) {
                        e = mp[k - C4(t) - cnt];
                        break;
                    }
                }
                if (e != -1)    break;
            }
            if (e != -1)    break;
        }
        if (e != -1)    break;
    }
    printf("%lld %lld\n", t + 5, t * (t - 1) / 2 + a + b + c + d + e);
    for (int i = 1; i <= t; i++) {
        for (int j = i + 1; j <= t; j++) {
            printf("%d %d\n", i, j);
        }
    }
    for (int j = 1; j <= a; j++)    printf("%lld %d\n", t + 1, j);
    for (int j = 1; j <= b; j++)    printf("%lld %d\n", t + 2, j);
    for (int j = 1; j <= c; j++)    printf("%lld %d\n", t + 3, j);
    for (int j = 1; j <= d; j++)    printf("%lld %d\n", t + 4, j);
    for (int j = 1; j <= e; j++)    printf("%lld %d\n", t + 5, j);
}