//函数fun功能:求n(n<10000)以内的所有四叶玫瑰数并逐个存放到result所指数组中,个数作为返回值.如果一个4位整数等于其各个位数字的4次方之和,则称该数为函数返回值. #include<stdio.h> #pragma warning (disable:4996) int fun(int n, int result[]) { ,j=; int a, b, c, d; ; i < n; i++) { a = i / ; b = (i % ) / ; c = (i %…
［抄题］: Given two strings s and t which consist of only lowercase letters. String t is generated by random shuffling string s and then add one more letter at a random position. Find the letter that was added in t. Example: Input: s = "abcd" t = &q…
<!DOCTYPE html><html><head> <title></title></head><script type="text/javascript"> function search(str1,str2) { var i=j=k=a=jk=kk=0; var m=str1.length; var n=str2.length; var index=0; var maxlen=0; var st…
Long Long Message Time Limit: 4000MS Memory Limit: 131072K Total Submissions: 31904 Accepted: 12876 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days…
Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. Example 1: Input: s1 = "sea", s2 = "eat" Output: 231 Explanation: Deleting "s" from "sea" adds the ASCII value of…
Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. Example 1: Input: s1 = "sea", s2 = "eat" Output: 231 Explanation: Deleting "s" from "sea" adds the ASCII value of…
Freedom of Choice URAL - 1517 Background Before Albanian people could bear with the freedom of speech (this story is fully described in the problem "Freedom of speech"), another freedom - the freedom of choice - came down on them. In the near fu…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
//动态规划查找两个字符串最大子串 public static string lcs(string word1, string word2) { int max = 0; int index = 0; int[,] nums = new int[word1.Length + 1,word2.Length+1]; for (int i = 0; i <= word1.L…
给定两个字符串 s 和 t,它们只包含小写字母.字符串 t 由字符串 s 随机重排,然后在随机位置添加一个字母.请找出在 t 中被添加的字母. 示例: 输入: s = "abcd" t = "abcde" 输出: e 解题思路:该题的解法和上一篇我们解决问题的思路一样,同样此题我们需要定义两个数组arr1和arr2分别存储字符串s和t每一个字符出现的次数,遍历统计字符串每一个字符出现的次数,最后遍历,找出arr1和arr2不相等,不相等的坐标存储的就是字符串t中添加…
问题描述: 题目描述Edit DistanceGiven two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)You have the following 3 operations permitted on a word: a) Insert a character …
今天碰到一个算法题觉得比较有意思,研究后自己实现了出来,代码比较简单,如发现什么问题请指正.思路和代码如下: 基本思路:从左开始取str的最大子字符串,判断子字符串是否为str的后缀,如果是则返回str加子字符串剩余部分:如果不是则逐步减少子字符串长度后在进行比较./* * 给出一个字符串s,输出包含两个字符串s的最短字符串,如s为abca时,输出则为abcabca */ public class ContainTwoString { public static String MergeStri…
#两个字符串,s1 包含 s2,包含多次,返回每一个匹配到的索引 def findSubIndex(str1,subStr): str_len = len(str1) sub_len = len(subStr) index_list = [] for i in range(str_len): for k in range(sub_len): if str1[i+k] != subStr[k]: break if k >= sub_len-1: index_list.append(i) print…
