2311. Longest binary subsequence less than or equal to K

Ideas ：

1. From right to left , Look for lower punctuation greater than the maximum

2. If there are remaining characters , Record only , by 0 Characters of

class Solution {
public:
    int longestSubsequence(string s, int k) {
        int ans = 0;
        
        //  Record the current total message 
        long sum = 0;

        //  Traverse the elements in reverse order , It is convenient to calculate numbers and... Directly from the right 
        int len = s.size() - 1;
        int i = len;
        for(; i >= 0; i--){
          //  Current digit 2 The power of 
          int bit = len - i;
          
          if(s[i] == '0'){
            //  If the number represented by the current power , exceed k, You can go back to , Prevent the following number overflow problem 
            if(k < (sum + pow(2, bit))){
              break;
            }
            
            //  Otherwise continue 
            continue;
          }
          
          //  If the current character is 1, Then it means that the number can be accumulated 
          //  When the accumulated number  > k It's about jumping out of the loop , Current i Keep in this position 
          if(k < (sum += pow(2, bit))){
            break;
          }
        }
      
        //  The number represented by this string completely conforms to 
        if(i == -1){
          return s.size();
        }
      
        //  From right to left , Length already met ,
        ans = len - i;
      
        //  Because the number represented has reached the limit , Only zero characters are recorded here 
        for(; i >= 0; i--){
          if(s[i] == '1'){
            continue;
          }
          
          ans++;
        }
      
        return ans;
    }
};