PotD2.c

/*
Most comments are obvious; akward structure due to quick coding
Recursion would have made this much easier to follow

Concept:  A,BCD = A,000 + B00 + C0 + D

All numbers will occur in less significant digit based on only the 'letter'
. For every 10, each digit will occur once in the 1's spot;
. For every 100, each digit will occur once in the 1's spot for every 10 (adds up to 10), and will be in tens spot for 10; etc 
All += C * 1 * 10^1; All += B * 2 * 10^2; All += A * 3 * 10^3; etc.

numbers in most significant digit will also be incremented:
most significant is incremented as other digits would count down (e.g. 113 - 100 has 14 100s with other values)
[A] += BCD + 1; [B] += CD + 1; [C] += D + 1; (+1 due to zero index: A,000, B00, etc.)

numbers in range below most significant digit will also be incremented:
for full countdown (e.g. 399 - 300 has 100 '300s')
0 is not included because 0BCD, 00CD, etc are not written as such
[1,A) += 1,000; [1,B) += 100; [1,C) += 10; [1,D) += 1;
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int count[10]; //count each digit, all initialized to 0

main() {


    	int i;          //for iterations
	int digit = 1;  //location in string
	int weight;     //used in pattern and obtaining remainders
	char n[6];      //string
	scanf("%s", n); //read input  (This is only running once, need to add another loop)
	int value = atoi(n);        //retrieve value of string
	int a = strlen(n) - 1;      //last digit at this location
	int add = 0;                //will record initial pattern
	for (weight = 1; digit <= a; weight *= 10, digit++) {      //counts all but most significant digit; based on pattern
		add += (n[a - digit] - '0') * digit * weight;
	}

	for (i = 0; i < 10; ++i)                    //This consistent amount is added to every count
		count[i] += add;


	digit = 0;                      //will start over at most significant digit
	int current, remainder;         //current digit and remainder = number to right of current
	while (digit <= a) {            //this solves for count of most significant digit of every number
		current = n[digit] - '0';   //idea is: 214 -> 200 is 15 most significant 2's, followed by 100 1's as 199 -> 100
		remainder = value % weight + 1;
		count[current] += remainder;
		for (i = current - 1; i > 0; i--) {
			count[i] += weight;
		}
		digit++;        //adjust as significant digit changes (eg 10 -> 9)
		weight /= 10;
	}

	for (i = 0; i < 10; i++) {      //prints counts
		printf("%d ", count[i]);
	}
	printf("\n");

	return 0;
}