Reputation: 99
What I need to do:
A program where the user inputs a range of numbers from S to K. It should count the amount of times a digit is found among the numbers, and finally print them out in the following order: 0 1 2 3 4 5 6 7 8 9.
My code:
OBS: It's not working for digits over 10.
#include <iostream>
using namespace std;
int main()
{
int S, K;
cin >> S >> K;
int digits = 0;
int dig0 = 0;
int dig1 = 0;
int dig2 = 0;
int dig3 = 0;
int dig4 = 0;
int dig5 = 0;
int dig6 = 0;
int dig7 = 0;
int dig8 = 0;
int dig9 = 0;
int remainder = 0;
for (int i = S; i <= K; i++) {
if (i < 10) {
switch(i) {
case 0:
dig0++;
break;
case 1:
dig1++;
break;
case 2:
dig2++;
break;
case 3:
dig3++;
break;
case 4:
dig4++;
break;
case 5:
dig5++;
break;
case 6:
dig6++;
break;
case 7:
dig7++;
break;
case 8:
dig8++;
break;
case 9:
dig9++;
break;
}
} else if (i >= 10) {
while (i > 0) {
remainder = i % 10;
switch(remainder) {
case 0:
dig0++;
break;
case 1:
dig1++;
break;
case 2:
dig2++;
break;
case 3:
dig3++;
break;
case 4:
dig4++;
break;
case 5:
dig5++;
break;
case 6:
dig6++;
break;
case 7:
dig7++;
break;
case 8:
dig8++;
break;
case 9:
dig9++;
break;
}
i /= 10;
}
}
}
cout << dig0 << ' ' << dig1 << ' ' << dig2
<< ' ' << dig3 << ' ' << dig4 << ' ' << dig5
<< ' ' << dig6 << ' ' << dig7 << ' ' << dig8 << ' ' << dig9;
return 0;
}
Desired behaviour:
Test case 01
Input: 1 to 9
Output: 0 1 1 1 1 1 1 1 1 1 (Since there are 0 digits of 0, and 1 digit of each number from 1 to 9)
Test case 02
Input: 767 772
Output: 1 1 1 0 0 0 3 10 1 1
Upvotes: 1
Views: 896
Reputation: 32586
(edit to allow any values for S and K including negative)
Can be
#include <iostream>
using namespace std;
int main()
{
int S, K;
int dig[10] = { 0 };
if (!(cin >> S >> K))
return -1;
if (S > K)
swap(S, K);
for (auto i = S; i <= K; ++i) {
auto v = i;
do {
dig[abs(v % 10)] += 1;
v /= 10;
} while (v != 0);
}
for (auto v : dig)
cout << v << ' ';
cout << endl;
return 0;
}
I did some change from your proposal
if (!(cin >> S >> K)) ...
)int v = i; ... dig[abs(v % 10)] += 1;
rather than int v = abs(i); ... dig[v % 10] += 1;
to manage the case where v is INT_MIN
( -INT_MIN
equals INT_MIN
so is still negative when using complement to 2)Compilation and executions :
pi@raspberrypi:/tmp $ g++ -pedantic -Wextra m.cc
pi@raspberrypi:/tmp $ ./a.out
1 9
0 1 1 1 1 1 1 1 1 1
pi@raspberrypi:/tmp $ ./a.out
767 772
1 1 1 0 0 0 3 10 1 1
pi@raspberrypi:/tmp $ ./a.out
-1 1
1 2 0 0 0 0 0 0 0 0
pi@raspberrypi:/tmp $ ./a.out
-767 -772
1 1 1 0 0 0 3 10 1 1
Additional note : if the auto disturb you replace them by int, and for (auto v : dig) cout << v << ' ';
by for (int i = 0; i != 10; ++i) cout << dig[i] << ' ';
Upvotes: 2
Reputation: 206
The use of the the switch statement is awkward and the multiple int variables can be consolidated in an array.
#include <array>
#include <iostream>
int main() {
std::array<int, 10> digits = {0};
int s, k;
std::cin >> s >> k;
int temp;
if(s > k) {//swap s and k, if s > k
temp = s;
s = k;
k = temp;
}
for(s; s <= k; s++) {
temp = s;
do {
digits[temp%10]++;
temp /= 10;
}while(temp != 0);
}
for(int j : digits) {
std::cout << ' ' << j;
}
return 0;
}
Upvotes: 0