title: Dice Throw Problem
signature: diceThrowProblem(m, n, x) {
test_code: import 'dart:convert';
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(int actual, int expected){
      return actual.toString() == expected.toString();
  }
  
  String driver(m, n, x, int expected){
      try
      {
          if (validateSolution(PLACEHOLDER_FN_NAME(m, n, x), expected)){
              return "PASSED";
          }
          return "FAILED";
      }catch(exception_object){
          return exception_object.runtimeType.toString();
      }
  }
  
  int main() {
  
      print("TEST-0...${driver(
          94, 4, 69, 
          50116)
      }");
  
      print("TEST-1...${driver(
          7, 12, 33, 
          77635844)
      }");
  
      print("TEST-2...${driver(
          20, 44, 24, 
          0)
      }");
  
      print("TEST-3...${driver(
          90, 94, 88, 
          0)
      }");
  
      print("TEST-4...${driver(
          50, 58, 27, 
          0)
      }");
  
      print("TEST-5...${driver(
          32, 90, 29, 
          0)
      }");
  
      print("TEST-6...${driver(
          46, 25, 6, 
          0)
      }");
  
      print("TEST-7...${driver(
          43, 82, 70, 
          0)
      }");
  
      print("TEST-8...${driver(
          6, 83, 19, 
          0)
      }");
  
      return 0;
  }
source_py: def dice_throw_problem(m, n, x):
      table = [([0] * (x + 1)) for i in range((n + 1))]
      for j in range(1, min((m + 1), (x + 1))):
          table[1][j] = 1
      for i in range(2, (n + 1)):
          for j in range(1, (x + 1)):
              for k in range(1, min((m + 1), j)):
                  table[i][j] += table[(i - 1)][(j - k)]
      return table[(- 1)][(- 1)]
source_cpp: int dice_throw_problem(int m, int n, int x) {
    int table [ n + 1 ] [ x + 1 ];
    memset ( table, 0, sizeof ( table ) );
    for ( int j = 1;
    j <= m && j <= x;
    j ++ ) table [ 1 ] [ j ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) for ( int j = 1;
    j <= x;
    j ++ ) for ( int k = 1;
    k <= m && k < j;
    k ++ ) table [ i ] [ j ] += table [ i - 1 ] [ j - k ];
    return table [ n ] [ x ];
  }