Y-25 Number of every character

Description
The look and say sequence is defined as follows. Start with any string of digits as the first element in the sequence. Each subsequent element is defined from the previous one by "verbally" describing the previous element. For example, the string 122344111 can be described as "one 1, two 2's, one 3, two 4's, three 1's". Therefore, the element that comes after 122344111 in the sequence is 1122132431. Similarly, the string 101 comes after 1111111111. Notice that it is generally not possible to uniquely identify the previous element of a particular element. For example, a string of 112213243 1's also yields 1122132431 as the next element.
Input
The input consists of a number of cases. The first line gives the number of cases to follow. Each case consists of a line of up to 1000 digits.
Output
For each test case, print the string that follows the given string.
Sample Input

3
122344111
1111111111
12345

Sample Output

1122132431
101
1112131415
解题思路：
统计每个字符连续出现的个数，不要当字符改变时，对于新出现的字符重新统计，
定义string t;用t[0]来存放当前字符，如果循环时，循环到的字符与当前字符相同，
则另存放字符个数的整数加1，如果循环到的字符与当前字符不同，则输出，此
时出现新字符，另存放字符个数的整数赋值1，并更改当前字符，
细节处理：
对于循环到最后一个字符时，直接输出，然后当前循环结束
代码：
#include<iostream>
#include<queue>
#include<map>
#include<string>
using namespace std;
int main()
{
int n,i;
cin>>n;
for(i=0;i<n;i++)
{
string s,t;
cin>>s;
t=s[0];
int j,temp=0;
for(j=0;j<s.size();j++)
{
if(s[j]==t[0]) 
{
temp++;
if(j==s.size()-1) cout<<temp<<t[0];
}
else 
{
cout<<temp<<t[0];
temp=1;
t[0]=s[j];
if(j==s.size()-1)
cout<<temp<<t[0];
}
}
cout<<endl;
}
return 0;
}