Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36852    Accepted Submission(s): 16344

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

2
ABC
ABBCCC

Sample Output

ABC
A2B3C

注意子字符串定义：

substring"abc"的substring包括"ab","bc","a"等等

必须是连续的，如果不连续如"ac"，则叫做subsequence （子序列）

#include<iostream>
#include<string>//这里用cstring oj编译不通过

using namespace std;

int main()
{
int caseNum=0;
cin>>caseNum;
while(caseNum--)
{
string strs;
cin>>strs;
int len=strs.length();
for(int i=0;i<len;i++)
{
char ori=strs[i];
if((i-1>-1)&&(strs[i]==strs[i-1]))
{
if(i==len-1)cout<<endl;
continue;
}
int count=1;
for(int j=i+1;j<len;j++)
{
if(ori!=strs[j])
{
break;
}
else
{
count++;
}
}
if(count==1)
{
cout<<ori;
}
else
{
cout<<count<<ori;
}
if(i==len-1)cout<<endl;
}
}
return 0;
}