Program of c to find largest and second larhest number of given 50 numbers using array

That appears to be an intermediate question in a programming class. Best read the book, the answer is not what is important the process of getting the answer is what you are trying to learn.

Write a program to find out the largest number from a
given unordered array of 8-bit numbers. Store the largest
number in largest variable.

52h, 23h, 56h, 45h, 9Ah, ABh

coding in C++

#include<iostream.h>
#include<conio.h>
#define max 50

void main()
{
clrscr();
int arr[max],size=0;
int i=0,j=0,temp;
do
{
cout<<"Enter the size of array(less than or equal to 50)";
cin>>size;
} while(size>50);

for(i=0;i<size;i++){
cout<<"Enter a number";
cin>>arr[i];
}


for( i=1;i<size;i++){
for(j=0;j<size;j++){
if(arr[j+1]>arr[j]){
temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;

}
}
}
cout<<”Array in descending order”;
for(i=0;i<size;i++){
cout<<arr[i]<<"\n";
}
getch();
return 0;
}

Need help doing your homework?

Here's a solution, written in BASIC:

10 PRINT, "Hello World"
20 EXIT

http://www.cplusplus.com/doc/tutorial/

Write a program to prompt the user to input several lines of input. A line of inputs is terminated by '*'. For each readline of input the program should output:(use pointer notation for the array) . The total # of words . The lenght of the longest word in the line At the end the program should output: . The total number of words read in all lines of input . The lenght of the longest word read in all lines of input

To start points and lines are not shapes, and a triangle is not representable with a single function, therefore you need a function to draw shapes , given a number of points as parameters, and one to draw math functions (lines etc.).

Here is an example: Draw Line, Ellipse, Polygon ...

From website above, here is script in JS:

"<script type="text/javascript">
<!--
function myDrawFunction()
{
jg_doc.setColor("#00ff00"); // green
jg_doc.fillEllipse(100, 200, 100, 180); // co-ordinates related to the document
jg_doc.setColor("maroon");
jg_doc.drawPolyline(new Array(50, 10, 120), new Array(10, 50, 70));
jg_doc.paint(); // draws, in this case, directly into the document

jg.setColor("#ff0000"); // red
jg.drawLine(10, 113, 220, 55); // co-ordinates related to "myCanvas"
jg.setColor("#0000ff"); // blue
jg.fillRect(110, 120, 30, 60);
jg.paint();

jg2.setColor("#0000ff"); // blue
jg2.drawEllipse(10, 50, 30, 100);
jg2.drawRect(400, 10, 100, 50);
jg2.paint();
}

var jg_doc = new jsGraphics(); // draw directly into document
var jg = new jsGraphics("myCanvas");
var jg2 = new jsGraphics("anotherCanvas");

myDrawFunction();

//-->
</script>"

This is a common starter challenge in college and high school programming classes. While i'm not willing to code it for you, I can show you the methods you'll need to understand in order to approach the problem. The easiest way of doing this is by using a brute force method. Just keep testing the number to see if the remainder after dividing it with a number lower than it is zero. If so, it's a factor.
For example, If the number is 35 then pick a number just one lower than it that's not the number 1. Now divide the two numbers. 35/34 = a number with many decimal places after it, which means its not one of the factors of the number. However, eventually, you'll approach the number 7, and 35/7 = 5. This number has no decimal, meaning it IS a factor of 35, and also the largest non-zero factor of 35. Your algorithm can now stop, since it found the largest non-zero number.
Here's some pseudo-code:
int largestfactor(int number) {
int temp = number;
for(temp - 1; temp>=1; temp--) {
if (number % temp == 0) return temp
}
return temp
}
Something like that, with some bug tweaks will do it. The "%" sign is called a modulus. It's a standard c++ operator. Good luck, and remember to vote Fixya if you liked it.
The wikipedia article that helps you with better and more effecient algorithms can be found here.
Steven

Haha, i wrote one of thee a while ago for a competition. You need to us e the char at function mate :D
Inside a for loop
Ill havea look for you though :D