Problem A. Ponding WaterDescriptionThere is a strange building on planet VOID. It is made up of N*Mrectangular parallelepipeds on N*M grids, whose bottom surfaces are
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2012-03-24 22:28:00
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Problem A. Ponding WaterDescriptionThere is a strange building on planet VOID. It is made up of N*Mrectangular parallelepipeds on N*M grids, whose bottom surfaces are 1*1squares. Since heights
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2023-09-15 09:44:54
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At the request of a user, you issue the following command to restore a dropped table: flashback table "BIN$F2JFfMq8Q5unbC0ceE9eJg==$0" to before drop;
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2017-11-13 15:39:00
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public class Solution {
public bool CanConstruct(string ransomNote, string magazine) {
Dictionary<char, int> dic = new Dictionary<char, int>();
//将ransomNote放入字典中
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2017-04-19 11:11:00
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Given an arbitrary ransom note string and another string containing letters from all the magazines, wri
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2022-08-03 16:54:12
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http://www.elijahqi.win/archives/645 C. Propagating treetime limit per test2 secondsmemory limit per test256 megaby
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2022-08-08 14:26:40
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题目：https://leetcode-cn.com/problems/ransom-note/ 想法：与”242.有效的字母异位词“一样 代码： C++版本： bool canConstruct(string ransomNote, string magazine) { int dictionar ...
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2021-10-20 10:33:00
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题目：
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magaz
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2023-03-07 12:36:17
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class Solution: def canConstruct(self, ransomNote: str, ma
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2022-07-01 13:51:03
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Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the rans
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2016-08-31 22:37:00
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Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the rans
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2020-05-06 10:37:00
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给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。 (题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表 ...
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2021-09-19 14:23:00
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给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。 (题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表 ...
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2021-09-25 22:00:00
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383. Ransom Note*https://leetcode.com/problems/ransom-note/题目描述Given an arbitrary ransom note string and another stri
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2022-05-30 10:48:36
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题目传送门#include <bits/stdc++.h>using namespace std;const int N = 1010;const int M = 10010;int n, m; // n个节点，m条边int d[N][2]; //拆点，因为每个点都要计算最短路和次短路，所以一个点看到两个点int cnt[N][2]; //最短路和次短路的路线个数
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2022-03-29 15:17:36
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给你两个字符串：ransomNote 和 magazine ，判断 ransomNote 能不能由 magazine 里面的字符构成。如果可以，返回 true ；否则返回 false 。magazine 中的每个字符只能在 ransomNote 中使用一次。示例 1：输入：ransomNote = "a", magazine = "b"输出：false示例 2：输入：ransomNote = "a
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2022-12-27 12:47:00
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题目来源：383. 赎金信 给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。 (题目说明：为了不暴露赎金信字迹，要从杂志上搜索
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2021-07-13 14:45:22
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/*Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines
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2021-07-09 14:05:19
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Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the rans
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2017-09-29 10:58:00
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题解 对于一对方向相反的牛，它们之间的顺序是无关紧要的，因为一定有一头会看到对方被挤奶，所以只需要考虑方向相同的牛。对于向左的牛，从右到左挤奶，反之亦然。 代码 #include<bits/stdc++.h> #define int long long using namespace std; co ...
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2021-09-24 15:41:00
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