| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 9475 | Accepted: 5773 |
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
3122344111111111111112345
Sample Output
11221324311011112131415
Source
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问题链接：。
问题简述：（略）
问题分析：这个问题是首先输入测试例子数量t，然后输入t行数字串，将每一行的数字串转换为Look and Say数字串输出。
程序说明：
这里给出两个程序，一个是将字符串读入到字符数组中，然后进行将数字串转换为Look and Say数字串的处理；另外一个程序是一边读入字符串，一边处理，省去缓冲存储。
使用getchar()函数直接处理输入流的话，不使用多余的缓存，是一个好主意。
参考链接：。
AC通过的C语言程序（正解）如下：
/* POJ3438 ZOJ2886 Look and Say */#include#define MAXN 1024char r[MAXN * 2];int main(void){ int t, say; char look, c, *pt, temp[10], *pt2; scanf("%d", &t); getchar(); while(t--) { // Look and Say转换：一边读入一边转换 pt = r; look = getchar(); // 读入首字符 say = 1; while((c = getchar()) != '\n') { if(c == look) say++; else { sprintf(temp, "%d", say); pt2 = temp; while(*pt2) *pt++ = *pt2++; *pt++ = look; look = c; say = 1; } } sprintf(temp, "%d", say); pt2 = temp; while(*pt2) *pt++ = *pt2++; *pt++ = look; *pt = '\0'; // 输出结果 printf("%s\n", r); } return 0;}
另外一个AC通过的C语言程序如下：
/* POJ3438 ZOJ2886 Look and Say */#include#define MAXN 1024char s[MAXN];char r[MAXN * 2];int main(void){ int t, say; char look, *ps, *pt, temp[10], *pt2; scanf("%d", &t); while(t--) { // 读入字符串 scanf("%s", s); // Look and Say转换 ps = s; pt = r; look = *ps; say = 1; while(*(++ps)) { if(*ps == look) say++; else { sprintf(temp, "%d", say); pt2 = temp; while(*pt2) *pt++ = *pt2++; *pt++ = look; look = *ps; say = 1; } } sprintf(temp, "%d", say); pt2 = temp; while(*pt2) *pt++ = *pt2++; *pt++ = look; *pt = '\0'; // 输出结果 printf("%s\n", r); } return 0;}