hdu 1020 简单的字符串处理

最新推荐文章于 2017-03-04 10:57:05 发布

转载最新推荐文章于 2017-03-04 10:57:05 发布 · 114 阅读

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原文链接：http://www.cnblogs.com/zn505119020/p/3582002.html

文章标签：

#java #php

本文介绍了如何使用特定方法对只包含大写字母的字符串进行编码，包括识别重复字符子串并压缩的过程。通过实例演示了从输入到输出的转换流程，并提供了实现该编码功能的源代码。

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1020

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23937 Accepted Submission(s): 10515

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

ABC

ABBCCC

Sample Output

ABC

A2B3C

代码如下：

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string>
 4 #include<string.h>
 5 #include<map>
 6 #include<math.h>
 7 #include<algorithm>
 8 #define N 10005
 9 using namespace std;
10 char str[N];
11 int main()
12 {
13     int n,i,t;
14     scanf("%d",&n);
15     while(n--)
16     {
17         scanf("%s",str);
18         i=0;
19         while(str[i]!='\0')
20         {
21             t=i;
22             while(str[t+1] == str[t])
23                 t++;
24             if(t>i)
25                 printf("%d%c",t-i+1,str[t]);
26             else
27                 printf("%c",str[t]);
28             i=t+1;
29         }
30         printf("\n");
31     }
32     return 0 ;
33 }