We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A. (Here, the given index = q…

package y2019.Algorithm.array; /** * @ProjectName: cutter-point * @Package: y2019.Algorithm.array * @ClassName: SumEvenAfterQueries * @Author: xiaof * @Description: 985. Sum of Even Numbers After Queries * * We have an array A of integers, and an arr…

problem 985. Sum of Even Numbers After Queries class Solution { public: vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) { vector<int> res; ; ==) sum +=a; for(auto query:queries) { ]]%== )…

Difficulty: Easy Question We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index]. Then, the answer to the i-th query is the sum of the even values of A. (…

We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A. (Here, the given index = q…

We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index]. Then, the answer to the i-th query is the sum of the even values of A. (Here, the given index = qu…

https://leetcode.com/problems/sum-of-even-numbers-after-queries/ We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th que…

题目要求 We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A. (Here, the given inde…

We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A. (Here, the given index = q…

这是悦乐书的第370次更新,第398篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第232题(顺位题号是985).有一个整数数组A和一个查询数组queries. 对于第i个查询val = queries[i][0],index = queries[i][1],我们将val添加到A[index].然后,第i个查询的答案是A的偶数值的总和.(这里给定的index = queries[i][1]是一个基于0的索引,每个查询都会修改数组A.) 返回所有查询的答案.你的答案数…

题目如下: We have an array A of integers, and an array queries of queries. For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A. (Here, the given ind…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力 找规律 日期 题目地址:https://leetcode.com/problems/sum-of-even-numbers-after-queries/ 题目描述 We have an array A of integers, and an array queries of queries. For the i-th query val = q…

Given a list of integers, write a function that returns the largest sum of non-adjacent numbers. Numbers can be 0 or negative. For example, [2, 4, 6, 2, 5] should return 13, since we pick 2, 6, and 5. [5, 1, 1, 5] should return 10, since we pick 5 an…

Check Sum of Square Numbers Given a integer c, your task is to decide whether there're two integers a and b such that a^2 + b^2 = c. 您在真实的面试中是否遇到过这个题? Yes 样例 Given n = 5Return true // 1 * 1 + 2 * 2 = 5 Given n = -5Return false 代码 class Solution { pub…

problem 633. Sum of Square Numbers 题意: solution1: 可以从c的平方根,注意即使c不是平方数,也会返回一个整型数.然后我们判断如果 i*i 等于c,说明c就是个平方数,只要再凑个0,就是两个平方数之和,返回 true:如果不等于的话,那么算出差值 c - i*i,如果这个差值也是平方数的话,返回 true.遍历结束后返回 false, class Solution { public: bool judgeSquareSum(int c) { ; --…

Two Sum 题目:https://leetcode.com/problems/two-sum/ class Solution(object): def twoSum(self, nums, target): map = {} for index, value in enumerate(nums): if target - value in map: return [map[target - value] + 1, index + 1] map[value] = index Add Two N…

由于深深的知道自己是事件驱动型的人,一直想补强自己的薄弱环节算法,却完全不知道从哪里入手.所以只能采用最笨的办法,刷题.从刷题中遇到问题就解决问题,最后可能多多少少也能提高一下自己的渣算法吧. 暂时的目标是一周最少两道,可能会多做多想,工作再忙也会完成这个最低目标. Two sum: Given an array of integers, return indices of the two numbers such that they add up to a specific target. Y…

题目连接:http://leetcode.com/2010/09/print-all-combinations-of-number-as-sum.html 题目分析: 由于这里说明了输入是升序的,当然如果是乱序的输入,只要没有要求输出有序,就不需要排序,否则在计算时,先对数组进行排序处理. 假设当前的arr[i]比Sum小,则计入arr[i],并更新Sum的值.否则,跳过当前的arr[i]元素. 题目扩展和变形: 假设这里的数字是不能重复的,参见不重复求和. 参考代码: void Solve(c…

Given a non-negative integer c, your task is to decide whether there're two integers a and bsuch that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False Accepted 38,694 Submissions 117,992  …

Difficulty: Easy  More:[目录]LeetCode Java实现 Description https://leetcode.com/problems/sum-of-square-numbers/submissions/ Given a non-negative integer c, your task is to decide whether there're two integers aand b such that a2 + b2 = c. Example 1: Inpu…

题意:给定一个非负整数c,确定是否存在a和b使得a*a+b*b=c. class Solution { typedef long long LL; public: bool judgeSquareSum(int c) { LL head = 0; LL tail = (LL)(sqrt(c)); while(head <= tail){ LL sum = head * head + tail * tail; if(sum == (LL)c) return true; else if(sum >…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3885 访问. 给定一个非负整数 c ,你要判断是否存在两个整数 a 和 b,使得 a2 + b2 = c. 输入: 5 输出: True 解释: 1 * 1 + 2 * 2 = 5 输入: 3 输出: False Given a non-negative integer c, your task is to decide whether there're two…

应该是纯模拟吧. 直接输入一个字符串,然后一位一位看,如果不是0,就 k++,并计算这个数的真实的值,最后输出就行了. #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define line cout << endl using namespace std; int t, k, sum = 1/*位数*…

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False 这道题让我们求一个数是否能由平方数之和组成,刚开始博主没仔细看题,没有看…

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5  Example 2: Input: 3 Output: False 给定一个非负整数 c ,你要判断是否存在两个整数 a 和 b,使得 a…

这是悦乐书的第276次更新,第292篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第144题(顺位题号是633).给定一个非负整数c,判断是否存在两个整数a和b,使得a的平方与b的平方之和等于c.例如: 输入:5 输出:true 说明:1 x 1 + 2 x 2 = 5 输入:3 输出:false 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法 暴力解法,直接使用两层for…

https://leetcode.com/problems/sum-of-square-numbers/ Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input:…

题目: Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False 分析: 给定一个非负整数c ,你要判断是否存在两个整数a和b,使…

static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { public: bool judgeSquareSum(int c) { ,b=sqrt(c); while(a<=b) { int k=a*a+b*b; if(k==c) return true; else if(k>c) b--; else a++; } return false; } }; 数学方法…

［抄题］: Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False [暴力解法]: 时间分析: 空间分析:n^2 [优化后]:…