HDU-1020-Encoding

最新推荐文章于 2024-07-21 10:00:00 发布

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最新推荐文章于 2024-07-21 10:00:00 发布

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原文链接：http://www.cnblogs.com/xiaxiaosheng/archive/2013/04/25/3043585.html

本文介绍了一种简单的字符串处理算法，该算法可以将由大写字母组成的字符串进行有效的压缩编码。通过对相同字符连续出现的情况进行计数，实现字符串的简化表示。文章通过一个具体的编码实例展示了算法的工作原理，并提供了一份C语言实现的源代码。

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

解题报告：字符串处理题

View Code

 1 #include<stdio.h>
 2 #include<string.h>
 3 char str[10005];
 4 int num[30];
 5 int main() {
 6     int n;
 7     scanf("%d",&n);
 8     while(n--) {
 9         scanf("%s",str);
10         int len=strlen(str);
11         int sum=1;
12         for(int i=0;i<len;++i)
13         if(str[i]==str[i+1])
14         sum++;
15         else {
16             if(sum==1)
17             printf("%c",str[i]);
18             else
19             printf("%d%c",sum,str[i]);
20             sum=1;
21         }
22         str[0]='\0';
23         printf("\n");
24     }
25     return 0;
26 }