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.

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.

For each test case, print the string that follows the given string.

3122344111111111111112345

11221324311011112131415

和1016题意一样，而且更简单。

题目要求将数n按一定规则进行转换，比如：1231560，这个数字中有1个0、2个1、1个2、1个3、1个5、1个6，则，其转换后的下一个数字为102112131516。

直接读取当前数字id，并同时计算个数num，当读取的数字和id不相同，则输出id和num，再将当前数字保存到id，知道字符串读取完成。

注意点：

1）末尾的id和num，可能忘记输出，造成错误。

代码（1AC）：

#include 
    
     #include 
     
      #include 
      
       char str[1100];int main(void){    int ii, casenum;    int len, num, id;    int i, j;    char last;    scanf("%d", &casenum);    getchar();    for (ii = 0; ii < casenum; ii++){        scanf("%s", str);        len = strlen(str);        for (i = 1, id = (str[0] - '0'), num = 1; i < len ; i++){            if (str[i] - '0' == id){                num++;            }            else{                printf("%d%d", num, id);                id = str[i] - '0';                num = 1;            }        }        printf("%d%d", num, id);        printf("\n");    }    return 0;}