hdu 1020 Encoding

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import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();while (n-- > 0) {String s = sc.next();String result = "";for (int i = 0; i < s.length(); i++) {int temp = repeat(s, i);if (temp == 1) {result += s.charAt(i);} else if (temp > 1) {result += temp;result += s.charAt(i);i += temp - 1;}}System.out.println(result);}}public static int repeat(String str, int start) {// 判断重复次数int count = 0;for (int i = start; i < str.length(); i++) {if (str.charAt(i) != str.charAt(start)) {return count;}count++;}return count;}}

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32155 Accepted Submission(s): 14272
编码

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:
给一个只包含A-Z的字符串。我们可以使用下列编码方法：
1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.
字符x表示A-Z内的字符，k表示子串中 x 出现的次数，输出是应该是kx。
2. If the length of the sub-string is 1, '1' should be ignored.

如果x只有一个那么k就不要表示。

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases.

输入的第一行是一个整数N，他表示测试事件的个数。

The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.

然后输入N行字符串。每一个字符都在A-Z之间且长度不大于1万。

Output

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

对于每一个测试事件，都在一行输出编码后的字符串。

Sample Input

2ABCABBCCC

Sample Output

ABCA2B3C

Author

ZHANG Zheng

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