< >

Keypad Count

TIME LIMIT = 1 SEC.

Okay, this problem has no story. A straightforward problem for your reading pleasure. Given N phone numbers, count the number of times each key has to be pressed to obtain each phone number.

Input	Output
The first line will contain N - the number of phone numbers. The following N lines will contain the N different phone numbers.	Ten space-separated integers, indicating the number of times each key from 0−9 was pressed for every phone number on a new line.

Constraints

1 ≤ N ≤ 100
0000000000 ≤ Phone Number ≤ 9999999999

Example Test Case

Input	Output
10 1508748728 8311124221 7960991616 2249409666 7385194952 3726036749 8963370335 6800221139 9407430513 4794136898	1 1 1 0 1 1 0 2 3 0 0 4 3 1 1 0 0 0 1 0 1 2 0 0 0 0 3 1 0 3 1 0 2 0 2 0 3 0 0 2 0 1 1 1 1 2 0 1 1 2 1 0 1 2 1 0 2 2 0 1 1 0 0 4 0 1 1 1 1 1 2 2 2 1 0 0 1 0 1 1 2 1 0 2 2 1 0 1 0 1 0 1 0 1 2 0 1 1 2 2

Solution

#include <iostream> #include <string> using namespace std; string counter(string ip) { int i,count[10]; string res; //resultant string for(i=0;i<10;i++) count[i]=0; //setting all digits count to zero for(i=0;i<10;i++) count[((int)ip[i])-48]++; //incrementing the value of the digit's count. //creating the result string by appending the count of each digit to the ampty string res. for(i=0;i<10;i++) res += to_string(count[i])+" "; return res; //returning the result string } int main() { int n,i; cin>>n; //number of phone numbers string ip[n]; // array of phone numbers stored as a string. for(i=0;i<n;i++) //taking in each phone number cin>>ip[i]; for(i=0;i<n;i++) cout<<counter(ip[i])<<endl; //printing the result for //each digit in the phone number using the // counter function defined above. return 0; }