xtu read problem training 2 B - In 7-bit

In 7-bit

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on  ZJU. Original ID: 3713
64-bit integer IO format: %lld      Java class name: Main

 

Very often, especially in programming contests, we treat a sequence of non-whitespace characters as a string. But sometimes, a string may contain whitespace characters or even be empty. We can have such strings quoted and escaped to handle these cases. However, a different approach is putting the length of the string before it. As most strings are short in practice, it would be a waste of space to encode the length as a 64-bit unsigned integer or add a extra separator between the length and the string. That's why a 7-bit encoded integer is introduced here.

To store the string length by 7-bit encoding, we should regard the length as a binary integer. It should be written out by seven bits at a time, starting with the seven least-significant (i.e. 7 rightmost) bits. The highest (i.e. leftmost) bit of a byte indicates whether there are more bytes to be written after this one. If the integer fits in seven bits, it takes only one byte of space. If the integer does not fit in seven bits, the highest bit is set to 1 on the first byte and written out. The integer is then shifted by seven bits and the next byte is written. This process is repeated until the entire integer has been written.

With the help of 7-bit encoded integer, we can store each string as a length-prefixed string by concatenating its 7-bit encoded length and its raw content (i.e. the original string).

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases.

Each test case is simply a string in a single line with at most 3000000 characters.

Output

For each test case, output the corresponding length-prefixed string in uppercase hexadecimal. See sample for more details.

Sample Input

3

42

yukkuri shiteitte ne!!!

https://en.wikipedia.org/wiki/Answer_to_Life,_the_Universe,_and_Everything#Answer_to_the_Ultimate_Question_of_Life.2C_the_Universe_and_Everything_.2842.29

Sample Output

023432

1779756B6B75726920736869746569747465206E65212121

9A0168747470733A2F2F656E2E77696B6970656469612E6F72672F77696B692F416E737765725F746F5F4C6966652C5F7468655F556E6976657273652C5F616E645F45766572797468696E6723416E737765725F746F5F7468655F556C74696D6174655F5175657374696F6E5F6F665F4C6966652E32435F7468655F556E6976657273655F616E645F45766572797468696E675F2E323834322E3239

 

Source

The 10th Zhejiang Provincial Collegiate Programming Contest

Author

WU, Zejun

 

解题：首先输出的字符串的长度编码。对的，每次输出7个二进制位，但是要用十六进制表示这7个二进制位表示的数字。把长度表示成二进制后，如果分割成一段7位二进制后，后面还有数字，那么，就要在输出的7个二进制位前面输出个1，在这七个二进制位前面输出一个0.后面的就是输出字符，把每个字符以十六进制数的形式输出即可。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #define LL long long

13 #define INF 0x3f3f3f3f

14 using namespace std;

15 char str[3000100];

16 int main(){

17     int t,i,len;

18     scanf("%d",&t);

19     getchar();

20     while(t--){

21         gets(str);

22         len = strlen(str);

23         if(!len){puts("00");continue;}

24         while(len){

25             int temp = len&127;

26             len >>= 7;

27             if(len) temp |= 128;

28             if(temp < 16) printf("0");

29             printf("%X",temp);

30         }

31         for(i = 0; str[i]; i++)

32             printf("%X",str[i]);

33         printf("\n");

34     }

35     return 0;

36 }

View Code