// Chain Explosions // Solution by Jacob Plachta #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; i=0; i--) #define FoxR1(i,n) for (i=n; i>0; i--) #define FoxRI(i,a,b) for (i=b; i>=a; i--) #define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++) #define Min(a,b) a=min(a,b) #define Max(a,b) a=max(a,b) #define Sz(s) int((s).size()) #define All(s) (s).begin(),(s).end() #define Fill(s,v) memset(s,v,sizeof(s)) #define pb push_back #define mp make_pair #define x first #define y second template T Abs(T x) { return(x < 0 ? -x : x); } template T Sqr(T x) { return(x * x); } string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); } const int INF = (int)1e9; const LD EPS = 1e-12; const LD PI = acos(-1.0); #define GETCHAR getchar_unlocked bool Read(int& x) { char c, r = 0, n = 0; x = 0; for (;;) { c = GETCHAR(); if ((c < 0) && (!r)) return(0); if ((c == '-') && (!r)) n = 1; else if ((c >= '0') && (c <= '9')) x = x * 10 + c - '0', r = 1; else if (r) break; } if (n) x = -x; return(1); } void ProcessCase() { // input int K; Read(K), K /= 2; // construct graph int N = 2; vector E = { { 1, 2 } }; while (K) { int r = N, d = 1; while (d <= K) E.pb({ r, ++N }), K -= d++; } // output printf("%d %d\n", N, Sz(E)); for (auto e : E) printf("%d %d\n", e.x, e.y); } int main() { int T, t; Read(T); Fox1(t, T) { printf("Case #%d: ", t); ProcessCase(); } return(0); }