// Perimetric - Chapter 1 // 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); } #define LIM 1000001 #define MOD 1000000007 int W; LL cur, ans; int L[LIM], H[LIM]; PR Q[LIM]; int Qs, Qe; void ReadSeq(int* V, int N, int K) { int i, A, B, C, D; Fox(i, K) Read(V[i]); Read(A), Read(B), Read(C), Read(D); FoxI(i, K, N - 1) V[i] = ((LL)A * V[i - 2] + (LL)B * V[i - 1] + C) % D + 1; } void ProcessRect(int L, int H) { // remove rectangles which are no longer relevant (too far left) while (Qe != Qs && Q[Qs].x < L) Qs++; if (Qe == Qs) cur += W + H; // no overlap, so add full rectangle perimeter else { cur += L + W - Q[Qe - 1].x; // add additional horizontal perimeter cur += max(0, H - Q[Qs].y); // add additional vertical perimeter } // remove rectangles which are no longer relevant (shorter than this one) while (Qe != Qs && Q[Qe - 1].y <= H) Qe--; // add this rectangle Q[Qe++] = mp(L + W, H); } int ProcessCase() { int N, K, i; // input Read(N), Read(K), Read(W); ReadSeq(L, N, K); ReadSeq(H, N, K); // process rectangles, while maintaining queue of relevant ones cur = 0, ans = 1; Qs = Qe = 0; Fox(i, N) { ProcessRect(L[i], H[i]); ans = ans * (cur % MOD) * 2 % MOD; } return(ans); } int main() { int T, t; Read(T); Fox1(t, T) printf("Case #%d: %d\n", t, ProcessCase()); return(0); }