UVa213
Message Decoding
题目虽然很长，但是很好理解。关键在于如何根据每个测试用例的第一行，构建解码表。
编码过程中，0和00代表不用的含义，所以直接用整数来索引解码表行不通，因此将解码表改为二维的，第一维表示长度，这样讲就能区分0和00了。
#include <iostream> #include <string> #include <cstring> using namespace std; void decode(const string &strHdr, const string &strBin) { char chDecode[7][128]; int len = 1, i = 0; memset(chDecode, 0, 7 * 128 * sizeof(char)); for (size_t idx = 0; idx < strHdr.size();idx++) { chDecode[len - 1][i] = strHdr[idx]; i++; if (i == ((1 << len) - 1)){ len++; i = 0; } } size_t curr = 0; string strMsg; while (1){ int len = stol(strBin.substr(curr, 3), nullptr, 2); curr += 3; if (len == 0) break; while (1){ int i = stol(strBin.substr(curr, len), nullptr, 2); curr += len; if (chDecode[len - 1][i] == '\0') break; else strMsg.push_back(chDecode[len - 1][i]); } } cout << strMsg << endl; return; } int main() { string strHdr; while (getline(cin, strHdr)){ string strLine, strBin; char ch; while (ch = cin.peek()){ if (ch != '0' && ch != '1') break; getline(cin, strLine); strBin += strLine; } decode(strHdr, strBin); } return 0; } /* TNM AEIOU 0010101100011 1010001001110110011 11000 $#**\ 0100000101101100011100101000 */