// Pond Precipitation // 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 32 #define LIM2 500 #define MOD 1000000007 int N; int D[LIM], C[LIM]; int ch[LIM2][LIM2]; int dyn[LIM][LIM2]; int Exp(int a, int b) { LL p = a, v = 1; while (b) { if (b & 1) v = v * p % MOD; p = p * p % MOD; b >>= 1; } return(v); } void Add(int& a, int b) { a = (a + b) % MOD; } int Prod(int a, int b) { return((LL)a * b % MOD); } int Div(int a, int b) { return Prod(a, Exp(b, MOD - 2)); } int ProcessCase() { int i, j, k; // input Read(N); Fox(i, N) Read(D[i]); // compute max prefix capacities int p = 0, s = 0; Fox(i, N + 1) { if (i == N || D[i] < D[p]) { FoxI(j, p, i - 1) s += D[j] - D[p]; FoxI(j, p + (p ? 1 : 0), i) C[j] = s; p = i; } } // DP Fill(dyn, 0); dyn[0][0] = 1; Fox(i, N) { Fox(j, (i ? C[i - 1] : 0) + 1) { Fox(k, C[i] - j + 1) Add(dyn[i + 1][j + k], Prod(dyn[i][j], ch[j + k][k])); } } // compute answer int ans = 0; Fox(i, s + 1) Add(ans, Div(dyn[N][i], Exp(N, i))); return(ans); } int main() { int T, t; int i, j; // precompute choose table ch[0][0] = 1; Fox1(i, LIM2 - 1) { Fox(j, i + 1) { ch[i][j] = ch[i - 1][j]; if (j) Add(ch[i][j], ch[i - 1][j - 1]); } } // testcase loop Read(T); Fox1(t, T) printf("Case #%d: %d\n", t, ProcessCase()); return(0); }