HDU 1020 Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45618    Accepted Submission(s): 20167

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

2ABCABBCCC

 

Sample Output

ABCA2B3C

 

              题意：

     给定一个仅包含‘A-Z’的字符串，用以下方法对它进行编码：

     1、包含K个相同字符的应编码为'KX',其中X是此子串中的唯一字符。

     2、如果子串长度为1，这忽略1。

     解题思路：

            找出子串中相同的字符，用一个计数器记录该字符出现的次数即可。若出现次数为1忽略1即可。

     注意：

     题目描述的是子串连续的的字符中各个字母的个数。而不是一整串的字符中的字母的个数

 例如：AABBBCAABCCD

       2A3BC2AB2CD

刚开始理解错题意了，错了好几遍呢！

代码：

#include <iostream>#include <algorithm>#include <map>#include <cstdio>#include <string>#include <cstring>using namespace std;int main(){    int t;    string s;    cin>>t;    getchar();    while(t--)    {        cin>>s;        int len=s.size();        int ans=1;///用ans来记录字符出现的次数        int temp=s[0];///用temp来存储第一个字符        for(int i=1; i<=len; i++)///将temp与后边的字符进行比较，相同的ans++        {            if(s[i]!=temp)            {                if(ans==1)///如果次数为1，忽略1，直接输出                {                    cout<<s[i-1];                    temp=s[i];                }                else                {                    cout<<ans<<s[i-1];                    temp=s[i];                    ans=1;                }            }            else                ans++;        }        cout<<endl;    }    return 0;}