Intermediate.
Put your skills to the test.

Problem summary

Give A…Z values 1…9 cyclically, ignoring case, and sum a name’s letter values. Repeatedly sum decimal digits until one digit remains, but stop immediately if the value is 33.

Constraints

The name contains at most 20 uppercase or lowercase Latin letters.

Worked example

Input and output for the worked example discussed below.

teodora

33

Solution analysis

1. The cyclic mapping is (uppercaseLetter−A) mod 9 + 1. Thus A,J,S all map to 1, while I,R map to 9. Sum these values before beginning decimal digit reduction.
2. While the total has at least two digits and is not 33, replace it by its decimal digit sum. Check the exception before every reduction, including the first one.
3. For teodora the mapped values are 2,5,6,4,6,9,1, summing to 33. The exception retains 33 rather than reducing it to 6. ISKRA instead sums to 22 and then becomes 4.

Why it works

The modulo expression implements exactly the repeated 1…9 letter table. Every reduction computes the decimal digit sum prescribed by the rules, and the loop stops exactly at one digit or the exceptional value 33.

Watch out. Do not use alphabet positions 1…26 without cycling, ignore case incorrectly, or reduce 33 to 6.

An independently written complete program with standard headers.

#include<cctype>
#include<iostream>
#include<string>
using namespace std;
int main(){
    string name;
    cin>>name;
    int key=0;
    for(char letter:name){
        int upper=toupper(static_cast<unsigned char>(letter));
        key+=(upper-'A')%9+1;
    }
    // The exceptional key must be checked before each reduction.
    while(key>9&&key!=33){
        int sum=0;
        while(key>0){
            sum+=key%10;
            key/=10;
        }
        key=sum;
    }
    cout<<key<<'\n';
}