HDOJ-1020 Encoding[水题]-优快云博客

本文介绍了一种简单的字符串压缩算法，该算法通过将连续重复的字符替换为字符出现次数及字符本身来实现。例如，字符串“ABCABBCCC”被压缩为“ABCA2B3C”。文章还提供了一个C++实现示例。

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16210 Accepted Submission(s): 6906

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, '1' should be ignored.

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.

Output

For each test case, output the encoded string in a line.

Sample Input

2 ABC ABBCCC

Sample Output

ABC A2B3C

Author

ZHANG Zheng

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JGShining

第一次提交WA了，后来搜了一下解题报告，这题没有要求按字典序排列。

code:

 1 #include<iostream>
 2 #include<string>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     int sum; 
 8     int i;
 9     char str[10002];
10     while(~scanf("%d",&n))
11     {
12         while(n--)
13         {
14             sum=0;
15             scanf("%s",&str);
16             for(i=0;i<strlen(str);i++)
17             {
18                 sum++;
19                 if(str[i]!=str[i+1])
20                 {
21                     if(sum==1)
22                          printf("%c",str[i]);
23                     else
24                         printf("%d%c",sum,str[i]);
25                     sum=0;
26                 }
27             } 
28             printf("\n"); 
29         }
30     }
31     return 0;
32 }