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Hi. I'm looking for a code to draw a random card for a small flash project i have due tomorrow. I've made 52 cards and given them all unique AS Linkage (Card0 - Card51). I just need a script that randomly chooses one of those cards.

I'm also willing to pay if i can get "live help" on creating this. I want to learn while doing it though which is why i'd rather get the answer and understand it instead of blindly pasting code.
cout << "Please enter ...";
cin >> numStrings;

string* words = new ________[ ___ ];

for (int i = 0; i < __________; _______)
{
getline(cin, words[ ____ ] );
[add a period to the end of the word]
}

int total = ____;

for (int n = 0; n < __________; _______)
{
int count = ____;
while ( _________[ ][ ] != '.')
{
count++;
}
cout << ________[ ] << " has " << ________ << "letters\n";
total = total + _____________;
}
cout << "Total number of letters is " << _________ << endl;
// main.cpp -- this program provides an example of how a program can **use** //
// the Stack Abstract Data Type (ADT). The stack adt's public functions are //
// **declared** in stack.h and **implemented** in stack.cpp. //
// //
// The purpose of main.cpp is to thoroughly test the functions of the stack //
// adt. //
// //
//

#include <iostream> // these two lines allow the program to use the
using namespace std; // standard cin and cout inpu
// stack.cpp implements a stack abstract data type. Users of the stack can //
// perform the following operations: //
// //
// 1. void initialiseStack(Stack) - reset the stack to the empty state //
// 2. bool isEmpty(Stack) - test whether the stack is empty //
// 3. bool isFull(Stack) - test whether the stack is full //
// 4. void push(Stack, StackElement) - put another element onto the stack //
// 5. StackElement pop(Stack) - remove & return the top element //
// 6. StackElement peek(Stack) - return a copy of the top element //
// //
// Internally, this stack abstract data type is implemented as an array of //
// elements.
// stack.h declares the public functions implemented in the stack abstract //
// (the implementation file is stack.cpp). By including this file in their //
// code, users of the stack can access the following operations: //
// //
// 1. void initialiseStack(Stack) - reset the stack to the empty state //
// 2. bool isEmpty(Stack) - test whether the stack is empty //
// 3. bool isFull(Stack) - test whether the stack is full //
// 4. void push(Stack, StackElement) - put another element onto the stack //
// 5. StackElement pop(Stack) - remove & return the top element //
// 6. StackElement peek(Stack) - return a copy of the top element //
// //
// The implementation of this stack can be found in
TASK C2.2: Explain why we wrote '==' rather than '=' in the expression.
// Make sure you explain what would happen if you write '=' rather
// than '==' in the expression



return (stack.top == BOTTOMOFSTACK ? true : false);
// TASK C2.1: explain what the expression between the parentheses below does.

return (stack.top == BOTTOMOFSTACK ? true : false);
// TASK C1.1: Explain the purpose of the '&' in a parameter.
// TASK C1.2: Explain what the program would do if the '&' was omitted in
// the initialiseStack() function header,
TASK H1: with each of the 4 marked statements immediately below, explain, in
// plain English, a) what the statement does and b) its purpose in the program.
const int MAXSTACKSIZE = 5;

// TASK H1.1 Explain this statement
const int BOTTOMOFSTACK = -1;

// TASK H1.2 Explain this statement
typedef char StackElement;
//
TASK H1.3 Explain this statemen

typedef struct {

// TASK H1.4 Explain this statement
StackElement contents[MAXSTACKSIZE];
int top;
} Stack;
Given a square maze (A) of dimension N, every entry (Aij) in the maze is either an open cell 'O' or a wall 'X'. A rat can travel to its adjacent locations (left, right, top and bottom), but to reach a cell, it must be open. Every cell in the maze will contain a survival probability P.

Your task is to find the number of ways to place maximum number of rats, such that they are not able to reach other and at the same time their minimum survival probability is at least PS. As the result can be long, take modulus of the result with 1000000007
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