HDU——1020 Encoding

本文主要是介绍HDU——1020 Encoding，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Encoding
*Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47730 Accepted Submission(s): 21182*

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 Input2
ABC
ABBCCC

Sample OutputABC
A2B3C

题目比较简单，思路就是以当前第一个与前面字符不相同的字符为基准，向后比较，找到第一个与它不同的字符。

C++代码：

#include <cstdio>
#include <cstring>
char str[10001];
//像这种比较大的数组定义在主函数外面比较好，有时候定义在里面会引起不必要的错误
int main(){int N,len=0,cur,com,temp;scanf("%d",&N);while(N-- != 0){scanf("%s",str);len = strlen(str);cur = 0;com = 0;while(cur < len && com < len){while(str[com] == str[cur] && com < len)com ++;temp = com - cur;if(temp > 1)printf("%d",temp);printf("%c",str[cur]);cur = com;}printf("\n");}
}

这篇关于HDU——1020 Encoding的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---