ACM--字母序列--HDOJ 1020--Encoding--字符串
来源:互联网 发布:党员干部必知的新词 编辑:程序博客网 时间:2024/06/06 07:14
HDOJ 题目地址:传送门
Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39606 Accepted Submission(s): 17517
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.
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
2ABCABBCCC
Sample Output
ABCA2B3C
题解:字符串打印的水题,模拟就可以了
#include<stdio.h>#include<iostream>#include<memory.h>#include<string.h>using namespace std;int main(){ int n; cin>>n; getchar(); while(n--){ string s; char temp; getline(cin,s); int index=1; temp=s[0]; for(int i=1;i<s.size();i++){ if(temp==s[i])//相同的自增 index++; else{//出现不同了 if(index==1){ printf("%c",temp); }else{ printf("%d%c",index,temp); } index=1; temp=s[i]; } } if(index==1) printf("%c\n",temp); else{ printf("%d%c\n",index,temp); } }}
1 0
- ACM--字母序列--HDOJ 1020--Encoding--字符串
- [HDOJ 1020]Encoding 字符串编码
- [HDOJ 1020]Encoding 字符串编码
- ACM--字母反转--HDOJ 1062--Text Reverse--字符串
- ACM--字母增加--HDOJ 1328--IBM Minus One--字符串
- ACM--字母排序--HDOJ 1379--DNA Sorting--字符串
- ACM--字母个数--HDOJ 1860--统计字符--字符串
- HDOJ 1020 Encoding
- hdoj 1020 Encoding
- hdoj 1020 Encoding
- HDOJ 1020 ENcoding
- hdoj 1020 encoding
- HDOJ 1020 Encoding
- HDOJ 1020 Encoding
- HDOJ 1020 Encoding
- hdoj-1020-Encoding
- HDOJ 1020 Encoding
- hdoj 1020 Encoding (水题)
- 利用IIS服务发布网站
- 我的wine中文字体配置笔记
- if else与三目运算符的区别
- iOS 获取网络图片的大小
- 二维数组排序 sort
- ACM--字母序列--HDOJ 1020--Encoding--字符串
- CF Round #361 (Div. 2) 689C. Mike and Chocolate Thieves
- Adb connection Error:远程主机强迫关闭了一个现有的连接
- 【2013国家队互测】家族(family)
- 菜鸟's problem
- 在android中如何用线程来更新UI
- Spring security防止跨站请求伪造(CSRF防护)
- web.py – xml模版
- 近红外摄像头Point-Grey开发日志