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// Chain Explosions | |
// Solution by Jacob Plachta | |
using namespace std; | |
template<typename T> T Abs(T x) { return(x < 0 ? -x : x); } | |
template<typename T> 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); | |
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<PR> 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); | |
} |