Compress the n-base array into a string (0-9 A-z) and decompress it.

There are two points to define such a question:

1. Create a table: index the character with an array subscript. Note that if the character is returned to a number:

Int Index = (STR [I]> = '0' & STR [I] <= '9 ')? (STR [I]-'0'-radix) :( STR [I]-'A'-radix + 10 );

2. Pay attention to whether the low position is in front of the high position. If the first position isLow * Radix ^ IYou can.

3. Calculate the number of Radix hexadecimal arrays in one bits...

This is mainly troublesome...

#include <assert.h>#include <algorithm>#include <vector>using namespace std;const int radix = 4;string base = "";string compress(const vector<int>& arr) {  string res = "";    int i, j, size = arr.size(), left, bits;    for (i = 1, j = 0; i*radix + radix < 36; i *= radix, ++j);  bits = j;  left = size - size / bits * bits;  size = size / bits * bits;  for (char ch = '0'; ch <= '9'; ++ch)    base.push_back(ch);  for (char ch = 'a'; ch <= 'z'; ++ch)    base.push_back(ch);  for (i = 0; i < size; i += bits) {    int index = 0, t = 1;    for (j = 0; j < bits; ++j) {      index = index + arr[i+j]*t;      t *= radix;    }               res.push_back(base[radix+index]);  }  for (i = 0; i < left; ++i)    res.push_back(arr[size+i]+'0');  return res; }vector<int> depress(const string& str) {  int len = str.length(), i = 0, j, bits;  for (i = 1, j = 0; i*radix + radix< 36; i *= radix, ++j);  bits = j;  vector<int> res;  for (i = 0; i < len; ++i) {    if (str[i]-'0' >= 0 && str[i]-'0' < radix)       res.push_back(str[i]-'0');    else {      int index = (str[i] >= '0' && str[i] <= '9') ? (str[i]-'0'-radix):(str[i]-'a'-radix + 10);            for (j = 0; j < bits; ++j) {        res.push_back(index%radix);        index = index/radix;      }    }  }  return res;} int main() {  int arr[] = {0,1,2,2,2,2,1,1,1,2,0,0,0,0,0,1};  vector<int> vec(arr, arr+sizeof(arr) / sizeof(int));  string str = compress(vec);  vector<int> res = depress(str);  return 0;}