Encoding Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 16   Accepted Submission(s) : 13 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the following meth…

A - Character Encoding HDU - 6397 思路 : 隔板法就是在n个元素间的(n-1)个空中插入k-1个板,可以把n个元素分成k组的方法 普通隔板法 求方程 x+y+z=10的正整数解的个数. 添元素隔板法 求方程 x+y+z=10的非负整数解的个数. 那么 增加 3 即转化为 了普通隔板法 但是这个题呢 还有 < N 的限制 ,那么就需要去除掉  ,分出的块中 有 > = n 的情况 . 就会 有 一块 出现 > =n ,两块 > =n 等等.. 具体…

Encoding Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32145 Accepted Submission(s): 14263 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the following me…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11132    Accepted Submission(s): 4673 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followin…

//1088(参考博客:http://blog.csdn.net/libin56842/article/details/8950688)//1.编写一个浏览器输入输出(hdu acm1088)://思路:对已经输入的字符串进行处理,遇到<br><hr>分别进行处理.遇到多于80个字符(统计该行的长度)或者<br>或者结束,则换行:遇到<hr>,输出80个'-'.#include<stdio.h>#include<string.h>#d…

pid=1020">Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25691    Accepted Submission(s): 11289 Problem Description Given a string containing only 'A' - 'Z', we could encode it usi…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39047    Accepted Submission(s): 17279 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51785    Accepted Submission(s): 23041 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

Problem Description In computer science, a character is a letter, a digit, a punctuation mark or some other similar symbol. Since computers can only process numbers, number codes are used to represent characters, which is known as character encoding.…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u 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 &quo…