I'm doing an online challenge and the challenge is the following: "Kids are playing a game called "Counting digits". For given numbers S and K, they firstly write all numbers between those numbers and then count how many times each digit appears (0,1,2,3,4,5,6,7,8,9). For example, S=767, K=772, numbers will be: 767,768,769,770,771,772
So, 0 will show once (in 770), 1 will show once (in 771) and so on..
Basically, my program have to do the following (given example):
Input: 1 9 (These are numbers 1,2,3,4,5,6,7,8,9)
Output: 0 1 1 1 1 1 1 1 1 1 (0 doesn't show, other numbers show once)."
I'm stuck on this code... out of ideas.
#include <iostream>
using namespace std;
int main()
{
int s,k;
int array[10];
int c0=0,c1=0,c2=0,c3=0,c4=0,c5=0,c6=0,c7=0,c8=0,c9=0;
cin >> s >> k;
int saves = s;
int savek = k;
cout << s%10;
for(int i=s;i<=k;i++)
{
int savei=i;
while(savei!=0)
{
savei=savei%10;
}
}
Any pseudo code/snippet/code/hint is appreciated.
Purely numeric solution to a purely numeric problem:
#include <iostream>
int main()
{
int s, k, i, tmp;
std::cin >> s >> k;
int count[10] = { 0 };
for (i = s; i <= k; i++) {
tmp = i;
do {
count[tmp % 10]++;
tmp /= 10;
} while(tmp);
}
for (i = 0; i < 10; i++) {
std::cout << i << " appears " << count[i] << " times" << std::endl;
}
return 0;
}
My solution is like this:
int main(){
int s,k;
cin >> s >> k;
int numbers[10]={0};
string sum;
for(int i=s;i<=k;i++)
{
sum=to_string(i);
for(int i=0;i<sum.length();i++){
numbers[(int)sum.at(i)-48]++;
}
}
for(int i=0;i<10;i++){
cout<<numbers[i]<<endl;
}
return 0;
}
public static void getDigitsInBook(int n) {
for(int i=0;i<10;i++) {
int x = n,val=0,k=1;
while(x!=0) {
int left = x/10;
int num = x%10;
int right = n%k;
if(i == 0) {
val = val+ (left*k);
}
else if(i<num) {
val = val + ((left+1)*k);
}
else if(i==num) {
val = val + (left*k) + right+1;
}
else {
val = val+ (left*k);
}
k=k*10;
x = n/k;
}
System.out.println(val);
}
}
What you usually do with such tasks is calculating the number between 0
and S
and between 0
and K
and subtracting those.
How many are between 0
and 767
? First count the numbers of the last digit. There are 77
times 0, 1, 2, 3, 4, 5, 6, 7 each and 76
times 8 and 9. More formally, 767/10+1
between 0
and 767%10
and 767/10+1
on the rest. Then calculate the number of occurences of the last digit for 767/10=76
, multiply by 10
, add 7
times 7 and 6 (for the error on the last one) and do the same for the remaining digits, here 76/10=7
. Finally, add the results up.
This solves the problem in O(log_10 K)
.
try this code:
for(int n=s ; n<=k ; n++)
{
tempN = abs(n);
while(tempN > 0)
{
tempDigit = tempN % 10;
tempN /= 10;
//count tempDigit here
}
}
assuming your variables are ints, "tempN /= 10;" should be no problem.
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