<input type="text" maxlength="25" oninput="textlength(this)"> <!--输入的内容--><span class="textNumber">0</span>个字符 <!--字符长度--> <!--调用的jquery方法--> function textlength(res) { var len =
Count and Say 思路:递归求出n - 1时的字符串,然后双指针算出每个字符的次数,拼接在结果后面 public String countAndSay(int n) { if(n == 1) return "1"; String front = countAndSay(n - 1); int i = 0; int j = 0; String res = ""; int count = 0; while(j < front.length()){ whi
题目链接:http://codeforces.com/problemset/problem/490/C 题目意思:给出一个可能有10^6 位长的字符串且没有前导0的整数,问能否一分为二,使得前面的一部分被 a 整除 且 后面那部分被 b 整除,可以的话输出 “YES” 和 划分后的两部分,否则输出“NO”. 看到字符串这么长,一下子就吓怕了---用 mod 即可!这些要靠数学积累吧...... 好常规的方法做——暴力枚举,当然不能遗漏啦,所以从左至右直到倒数第二个数字,都要算出mod a 的结果
Problem F "Folding" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 Description Bill is trying to compactly represent sequences of capital alphabetic characters from 'A' to 'Z' by folding repeating subsequences insid
str2int Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1568 Accepted Submission(s): 540 Problem Description In this problem, you are given several strings that contain only digits from '0'
Problem Description Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us. There is a large