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.
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
2 ABC ABBCCC
Sample Output
ABC A2B3C
Author
ZHANG Zheng