2020/finals/pond_precipitation.cpp

// Pond Precipitation
// Solution by Jacob Plachta

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <unordered_map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; 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<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);

#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);
}