Count a string from 1 instead of 0 in c++ - c++

Question

do you know how can I start to count a string from 1 and not from 0 in c++? for example, xyz are on iterations 0,1,2.. but i need them to be on 1,2,3. I know you can put cout << i+1; but is there any other way to do this with strlen(s+1)?
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char s[100];
cin.getline(s, 99);
int n = strlen(s);
for (int i = 0; i < n; ++i)
cout<<i<<": "<<s[i]<<"\n";
return 0;
}

Answer 1

i+1 is the way to go. An alternate method is starting your loop with 1 instead and accessing the array at element i-1, but that's a bit silly:
#include <iostream>
#include <cstring>
using namespace std;
int main() {
char s[100];
cin.getline(s, 99);
int n = strlen(s);
for (int i = 1; i <= n; ++i) // i = 1 not 0
cout<<i<<": "<<s[i-1]<<"\n";
return 0;
}

Answer 2

In C++, you shouldn't use C-Strings. Instead, use std::string:
#include <iostream>
#include <string>
int main()
{
std::string s;
std::getline(std::cin, s);
for (int i = 0; i < s.length(); ++i)
std::cout << i + 1 << ": " << s[i] << '\n' ;
return 0;
}

Answer 3

As others have said, the simple answer is really just accept you start from zero - however, for an actually solution you can use boost-range and use an index adaptor with an argument of '1'.
std::string str;
std::getline(std::cin, str);
for (const auto & element : str | boost::adaptors::indexed(1)) {
std::cout << element.index()
<< " : "
<< element.value()
<< std::endl;
}