Coin Change Problem

Given \(N\) denominations,how can a given amount of money \(V\) be made with the least number of coins? The most intuitive solution is greedy algorithm: sort the denominations, try the largest denominations lower than the value every time, and decrease the value with the denominations picked in every iteration. Repeat this process until the value reduces to 0. This greedy strategy works for the US (and most other) coin systems, However, it is not a general solution to all denomincations. For example, if the coin denominations were 1, 3 and 4, then to make 6, the greedy algorithm would choose three coins (4,1,1) whereas the optimal solution is two coins (3,3).

Text Justification

The Problem:

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C++ Code to Print Pascal Triangle

Printing Pascal Triangle seems like an easy problem, however, it is not that easy to print a good-looking Pascal Triangle.

To print the Pascal Triangle, for each line, first print spaces to the left of numbers, and then print digit numbers.

To calculate Pascal numbers, two assays can be used: one array to store numbers of above line, the other array to store numbers of this line. The first line has 1 number(1), the second lien has 2 numbers(1 1), the third line has three numbers(1 2 1) and so on.

To make the Pascal Triangle more readable, print spaces between two neighbor  numbers in the same line.

// Pascal triangle
#include <iostream>
#include <iomanip>
#include <cstring>
using namespace std;

const int WIDTH=7;
int main()
	int* oldarr=new int[WIDTH]; // store numbers of above row
	int* newarr=new int[WIDTH]; // store numbers of current row

	for(int i=0;i<WIDTH;++i) // cntrol row counting
		for(int j=0;j<WIDTH-i-1;++j) // left space
			cout << setw(2) << " ";

		// set default boundary numbers

			for(int k=1;k<i;++k)

		// Print the middle part
		int idx=0;
		int flag=(WIDTH-i-1)%2;
		for(int j=WIDTH-i-1;j<WIDTH+i;++j)
				cout << setw(2) << newarr[idx++];
			} else {
				cout << setw(2) << " ";
		//for(int j=WIDTH+i;j&lt;WIDTH;++j) // Print right space, if you like
		//	cout&lt;&lt;setw(4)&lt;&lt;&quot; &quot;;

		cout << endl;

	delete [] oldarr;
	delete [] newarr;

	return 0;

The output looks like:

           1   1
         1   2   1
       1   3   3   1
     1   4   6   4   1
   1   5  10  10   5   1
 1   6  15  20  15   6   1