Here Below I written my Solution of Digit Sum Parities in Two different Languages C++ and Python.
Solution in C++ :
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
ll digitsum(ll n)
{
ll ans = 0;
while (n)
{
ans += n % 10;
n /= 10;
}
return ans;
}
int main()
{
ll t = 1;
cin >> t;
while (t--)
{
ll n;
cin >> n;
ll sum = 0;
ll temp = n;
while (temp)
{
sum += temp % 10;
temp /= 10;
}
bool flag = true;
while (flag)
{
n += 1;
ll s = digitsum(n);
if (s % 2 != sum % 2)
{
cout << n << "\n";
flag = false;
break;
}
}
}
}
Solution in Python3 :
def digitsum(x):
sum = 0
while x != 0:
sum += x % 10
x = x // 10
return sum
t = int(input())
for _ in range(0,t):
n = int(input())
num1 = digitsum(n)
num2 = digitsum(n+1)
if (num1 % 2 != num2 % 2):
print(n+1)
else:
print(n+2)