1945. sum of digits after string conversion

1945. The sum of the numbers after string conversion

Give you a string of lowercase letters s , And an integer k .

First , Replace the letter with its position in the alphabet , take s conversion It's an integer （ That is to say ,‘a’ use 1 Replace ,‘b’ use 2 Replace ,… ‘z’ use 26 Replace ）. next , The integer transformation For its The sum of your numbers . Repeat transformation operation k Time .

for example , If s = “zbax” And k = 2 , Then the result obtained after performing the following steps is an integer 8 ：

 conversion ："zbax" * "(26)(2)(1)(24)" * "262124" * 262124
 transformation  #1：262124 * 2 + 6 + 2 + 1 + 2 + 4 * 17
 transformation  #2：17 * 1 + 7 * 8

Returns the result integer after performing the above operation .

Example 1：

Input ：s = “iiii”, k = 1
Output ：36
explain ： The operation is as follows ：

• conversion ：“iiii” * “(9)(9)(9)(9)” * “9999” * 9999
• transformation #1：9999 * 9 + 9 + 9 + 9 * 36
  therefore , The result integer is 36 .

Example 2：

Input ：s = “leetcode”, k = 2
Output ：6
explain ： The operation is as follows ：

• conversion ：“leetcode” * “(12)(5)(5)(20)(3)(15)(4)(5)” * “12552031545” * 12552031545
• transformation #1：12552031545 * 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 * 33
• transformation #2：33 * 3 + 3 * 6
  therefore , The result integer is 6 .

This problem is still very interesting. The solution code is as follows ：

int f(int num){
    
   int  sum=0;
     while(num){
    
                sum=sum+num%10;
                num=num/10;
                
            }
            return sum;

}
int getLucky(char * s, int k){
    
    int sum=0;
    int i;
    for(i=0;s[i]!='\0';i++){
    
        int num=s[i]-'a'+1;
        if(num<10){
    
            sum=sum+num;

        }
        else{
    
            while(num){
    
                sum=sum+num%10;
                num=num/10;
                
            }
            
        }

    }
    while(k>=2){
    
        sum=f(sum);
        k--;
    }
    return  sum;

}