吐槽下我的渣渣英语啊，即使叫谷歌翻译也没有看懂，最后还是自己读了好几遍题才读懂。

题目大意：题意很简单，就是给一些互不相同的由‘0‘，‘1‘组成的字符串，看看有没有一个字符串是否会成为另一个的开头的子串。

直接简单粗暴的去比较就可以了。

这是原题：

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols {A, B, C, D}

The following code is immediately decodable:

A:01 B:10 C:0010 D:0000

but this one is not:

A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)

Input 

Write a program that accepts as input a series of groups of records from a data file. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

Output 

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

The Sample Input describes the examples above.

Sample Input 

0110001000009011001000009

Sample Output 

Set 1 is immediately decodableSet 2 is not immediately decodable

AC代码：

 1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7  8 char code[9][11]; 9 bool cmp(char s1[], char s2[]);10 11 int main(void)12 {13     #ifdef LOCAL14         freopen("644in.txt", "r", stdin);15     #endif16     int kase = 0, n;17     while(gets(code[0]))18     {19         bool flag = false;20         int i, j;21         n = 0;22         while(code[n][0] != ‘9‘)23             gets(code[++n]);24 25         for(i = 0; i < n - 1; ++i)26         {27             if(flag)28                 break;29             for(j = i + 1; j < n; ++j)30             {31                 flag = cmp(code[i], code[j]);32                 if(flag)    break;33             }34         }35 36         if(!flag)37             printf("Set %d is immediately decodable\n", ++kase);38         else39             printf("Set %d is not immediately decodable\n", ++kase);40     }41     return 0;42 }43 //比较一个字符串是否会成为另一个的开头的子串44 bool cmp(char s1[], char s2[])45 {46     int l1 = strlen(s1);47     int l2 = strlen(s2);48     int lmin = min(l1, l2), i = 0;49     if(l1 == l2)//长度相等，必然不会是子串50         return false;51     for(i = 0; i < lmin; ++i)52     {53         if(s1[i] != s2[i])54             break;55     }56     if(i == lmin)57         return true;58     return false;59 }

代码君