杭电 1020 Encoding

这题不是很难，代码也不太简洁，其实可以把输出和对字符串的处理合并起来的，就更节省时间了。

/* THE PROGRAM IS MADE BY PYY */ /*----------------------------------------------------------------------------// http://acm.hdu.edu.cn/showproblem.php?pid=1020 Encoding Date : 2011/2/27 Thursday Begin : 10:57 End : 11:37 //----------------------------------------------------------------------------*/ #include <iostream> #include <string.h> using namespace std; char str[10001], res[27]; int cnt[28]; void process() { int j = -1; char prev_c = 0; for (int i = 0; str[i] != '/0'; i++) { // an old character if (prev_c == str[i]) { cnt[j]++; } // a new character else { res[++j] = str[i]; cnt[j] = 1; prev_c = str[i]; } } } void print(void) { for (int i = 0; res[i] != '/0'; i++) { if (cnt[i] == 1) { // do nothing } else cout << cnt[i]; cout << res[i]; } cout << endl; } int main() { int cnt[27], pcase; while (cin >> pcase) { while (pcase--) { memset(cnt, 0, sizeof(cnt)); cin >> str; process(); print(); } } return 0; }

======================================原题如下======================================

Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10131 Accepted Submission(s): 4172

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

Author
ZHANG Zheng

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