此类题目要明确两点：

1. 打表：用数组下标索引字符，同时注意如果从字符对应回数字： 

int index = (str[i] >= ‘0‘ && str[i] <= ‘9‘) ? (str[i]-‘0‘-radix):(str[i]-‘a‘-radix + 10);

2. 注意低位在前还是高位在前，如果先来的是 低位*radix^i 即可。

3. 统计每几个radix进制数组成一位，利用bits来表示。。。

这破题主要是麻烦。。。

#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;}