杭电 1020 Encoding

本文介绍了一种用于压缩字符串的简单编码算法,通过将连续重复的字符替换为字符及其出现次数来减少存储空间。该算法适用于仅包含A-Z的大写英文字母的字符串,并通过实例演示了如何实现这一算法。

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

Recommend
JGShining

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值