String count

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Title Description ：
Find the dictionary order in s1 and s2 Between , The length is in len1 To len2 Number of strings , result mod 1000007. Data range ：s1（ The length is less than 50）,s2（ The length is less than 50）,len1（ Less than 50）,len2（ Greater than len1, Less than 50） Be careful ： There are many groups of input
Input ：
Each set of data includes s1（ The length is less than 50）,s2（ The length is less than 50）,len1（ Less than 50）,len2（ Greater than len1, Less than 50）
ab ce 1 2
Output ：
56

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Ideas ：

1. Because it comes from the meaning of the title, it is a lowercase letter a~z, common 26 position , Think of it as 26 Binary operation
2. First step , Use append Complete the two strings to len2, It's easy to compare ,s1 Full complement ‘a’, s2 Full complement char(‘z’+1), Define another record 26 An array of hexadecimal digits arr, Subscript to be 0 The number of digits is the largest
3. i from 0 Start to len2 The current comparison between the two strings is the i position , Get the characters at the corresponding positions of the two strings in the loop arr Array （ Be careful ： Negative numbers need to borrow from the high position , The code here needs to be improved ）
4. i from len1 Start to len2, The number of representative digits is i A string that satisfies the conditions , for example arr={2,3,1}, The length is 1 The string of is 2*26⁰ The length is 2 The string of is 2*26¹+3*26⁰ The length is 3 The string of is 2*26²+3*26¹+1*26⁰
5. Note that the last string does not count result–

The code is as follows ：

int main() {
     
   string s1, s2;
   int len1, len2;
   while (cin >> s1 >> s2 >> len1 >> len2) {
     
       //  A string containing only lowercase letters can be regarded as 26 Hexadecimal number system 
       //  take s1 and s2 Lengthen to len2 length 
       s1.append(len2 - s1.size(), 'a');
       s2.append(len2 - s2.size(), (char)('z' + 1));
       
       //  confirm s1 and s2 The difference between the two strings at each position 
       vector<int> arr;
       for (int i = 0; i < len2; ++i)
       {
     
           if(s2[i] - s1[i]<0)
           {
     
               arr.back()-=1;
               arr.push_back(s2[i] - s1[i]+26);
           }
           else
               arr.push_back(s2[i] - s1[i]);
       }
       //  confirm len1 and len2 The number of different strings that can be formed between 
       int result = 0;
       for (int i = len1; i <= len2; ++i) 
           for (int k = 0; k < i; ++k)
               result += arr[k] * pow(26, i - 1 - k);
       //  All strings do not end with s2 Oneself , All the final reduction is 1
       cout << result  << endl;
   }
   return 0;
}