digit counter

[alejandro2]

hi everyone im new to c++ and sorry for my english.i want to make a basic program which count digits in number.after program starts we enter 5 number.after then program counts digits in number and shows digits used how many times in number.
dos screen must seem lik this:

enter integers between 1-5:
1
12
123
1234
12345 //we enter these numbers after program start

1:5 times
2:4 times
3:3 times
4:2 times
5:1 times

enter integers between 1-5:
_

i would be delighted if you help me.
thanks in advance:)

[firix]

Please share with us after you try to write code.

As an idea:

From 0 up to 9. Create an array of numbers.
You can count the number that you enter the digits.

[alejandro2]

i thought like this firix.but the program doesnt work.i dont understand my fault. i will be glad if you help me

#include <iostream.h>

#include <stdlib.h>

using namespace std;
main()

{

int array[10]={0,1,2,3,4,5,6,7,8,9};
int left=0,right=9,middle;
int number,i;
bool sw=false;


cout<<"ARRAY: ";

for (i=0;i<=9;i++)
{
cout<<array[i]<<";";
}
cout<<" \n"<<"Number to find: ";
cin>>number;
while (sw == false && left <= right)
{
middle=(left+right)/2;
if (number == array[middle])
{
sw=true;
cout<<"number found.\n";
}
else
{
if (number < array[middle]) right=middle-1;
if (number > array[middle]) left=middle+1;
}
}
if (sw == false) cout<<"number not found.\n";
system("Pause");
}

[firix]

int main()
{
	map<int, int> imap;
	int result;
	int value = 49549995;


	imap[0] = 0;
	imap[1] = 0;
	imap[3] = 0;
	imap[4] = 0;
	imap[5] = 0;
	imap[6] = 0;
	imap[7] = 0;
	imap[8] = 0;
	imap[9] = 0;

	while (value / 10) {
		result = value % 10;
		value /= 10;
		for (map<int, int>::iterator iter = imap.begin(); iter != imap.end(); iter++)
			if (iter->first == result)
				iter->second++;
	}


	for (map<int, int>::iterator iter = imap.begin(); iter != imap.end(); iter++)
		if(value == iter->first)
			iter->second++;

	

	for (map<int, int>::iterator iter = imap.begin(); iter != imap.end(); iter++)
		cout << iter->first << ":" << iter->second << endl; 
	


	return 0;
}

I hope this helps you.

[ne555]

If you are going to use a map:

/* traverse a map/set is O(n log n) (or at least O(n) )
for (map<int, int>::iterator iter = imap.begin(); iter != imap.end(); iter++)
			if (iter->first == result)
				iter->second++;*/
imap[result]++; // binary search O(log n)  

i want to make a basic program which count digits in number

But the code you post is binay search

[firix]

@ne555

please
you type in this code.

[Faff]

My try :D

#include <iostream>
using namespace std;

int main()
{
	
	int num[5]={0,0,0,0,0};
	int y=0;
	char x;

	do { //Loops while x is equal or higher then 1 and equal or lower then 5
		cin >> x;
/*		x == '1'?num[0]++:num[0]==num[0];//if x = 1 num[0]adds 1
		x == '2'?num[1]++:num[1]==num[1];//if x = 2 num[1]adds 1
		x == '3'?num[2]++:num[2]==num[2];//if x = 3 num[2]adds 1
		x == '4'?num[3]++:num[3]==num[3];//if x = 4 num[3]adds 1
		x == '5'?num[4]++:num[4]==num[4];//if x = 5 num[4]adds 1	
can obviously be simplified as ne555 says in the folowing post as:*/	
		num[ x-'1' ]++;
			}
	while ( x>='1',x<='5');	

	cout << "calculating ...\n";//Loop ended, displays the amount of each num occured
	cout << "1 occured " << num[0] << " times.\n";
	cout << "2 occured " << num[1] << " times.\n";
	cout << "3 occured " << num[2] << " times.\n";
	cout << "4 occured " << num[3] << " times.\n";
	cout << "5 occured " << num[4] << " times.\n";	

	cin.get();
	cin.get();//pause
return 0;
}

Hope this helped^^

btw if someone knows a alternative for the 0,0,0,0,0 in this line: int num[5]={0,0,0,0,0}; pls tell me :)
seems like I get a error if i didn't type it
kinda want to assign every variable in the array to 0 :)

[firix]

fantastic :)

[ne555]

too obfuscated
x == '1'? num[0]++: num[0]==num[0] /*statement has no effect */;

cin >> x;
num[ x-'1' ]++;

[Faff]

Thanks for pointing me to that ^^, would have took me a while to figure that out myselve :)
edited post with your code

[firix]

It does not matter ^^