若p=2或p=4*k+1 则p能够表成两平方数的和的形式 (欧拉和费马已证明,而且有求的方法) 所以答案是p 若p=4*k+3 设a^2=n(mod p) (n!=0)  能够证明不存在b,b^2=p-n(mod p) 即若n是p的平方剩余 则p-n不是p的平方剩余 证明:由于a^2=n(mod p) 所以由欧拉准则 得n^((p-1)/2)=1(mod p) 若b^2=-n(mod p) 那么(-n)^((p-1)/2)=1(mod p) 左边把符号提出来 得(-1)^((p-1)/2)*n^…

题目链接:1338: The minimum square sum Description Given a prime p(p<108), you are to find min{x2+y2}, where x and y belongs to positive integer, so that x2+y2=0 (mod p). 输入一个质数 p,你找出两个正整数 x 和 y 使得 (x2+y2) mod p = 0,且 x2+y2 最小. Input Every line is a p. No…

题目传送门 GCD SUM 题目描述 for i=1 to n for j=1 to n sum+=gcd(i,j) 给出n求sum. gcd(x,y)表示x,y的最大公约数. 输入输出格式 输入格式: n 输出格式: sum 输入输出样例 输入样例#1: 2 输出样例#1: 5 说明 数据范围 30% n<=3000 60% 7000<=n<=7100 100% n<=100000 分析: 无聊的出题人出的无聊的数学题. 这里博主用了一种比较暴力的思想,直接枚举以$1\thick…

Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time.   动态规划即可,与Unique…

题目描述: 题目链接:64 Minimum Path Sum 问题是要求在一个全为正整数的 m X n 的矩阵中, 取一条从左上为起点, 走到右下为重点的路径, (前进方向只能向左或者向右),求一条所经过元素和最小的一条路径. 其实,题目已经给出了提示:, 动态规划应该是最直接的解法之一. 这边我们了解到, 问题中只允许走到的每个点右移或者下移,这就意味着从起点开始, 都有两种后继路径(最后一行和最后一列除外),以此类推, 得到所有路径,然后取其中路径和虽小的值,就可以得到结果了. 但是!我们仔…

Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 还是DP问题,给定一个m*n的二…

之所以将这三道题放在一起,是因为这三道题非常类似. 1. Minimum Path Sum 题目链接 题目要求: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or righ…

Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. SOLUTION 1: 相当基础…

Unique Paths https://oj.leetcode.com/problems/unique-paths/ A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to rea…

Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 解题思路: 典型的动态规划.开辟…

引言 二维动态规划中最常见的是棋盘型二维动态规划. 即 func(i, j) 往往只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有关 这种情况下,时间复杂度 O(n*n),空间复杂度往往可以优化为O(n) 例题  1 Minimum Path Sum  Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whi…

Minimum Path Sum 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/minimum-path-sum/description/ Description Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its…

64. 最小路径和 64. Minimum Path Sum 题目描述 给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小. 说明: 每次只能向下或者向右移动一步. 每日一算法2019/5/23Day 20LeetCode64. Minimum Path Sum 示例: 输入: [ [1,3,1], [1,5,1], [4,2,1] ] 输出: 7 解释: 因为路径 1→3→1→1→1 的总和最小. Java 实现 略 相似题目 62. 不同路…

problem 599. Minimum Index Sum of Two Lists 题意:给出两个字符串数组,找到坐标位置之和最小的相同的字符串. 计算两个的坐标之和,如果与最小坐标和sum相同,那么将这个字符串加入结果res中,如果比sum小,那么sum更新为这个较小值,然后将结果res清空并加入这个字符串. solution: class Solution { public: vector<string> findRestaurant(vector<string>&…

Leetcode之动态规划(DP)专题-64. 最小路径和(Minimum Path Sum) 给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小. 说明:每次只能向下或者向右移动一步. 示例: 输入: [   [1,3,1], [1,5,1], [4,2,1] ] 输出: 7 解释: 因为路径 1→3→1→1→1 的总和最小. 找从左上角0,0到右下角的最短路径. DP:我们每个点(x,y)都可以表示为dp[x][y] = max( grid…

一.题目说明 题目64. Minimum Path Sum,给一个m*n矩阵,每个元素的值非负,计算从左上角到右下角的最小路径和.难度是Medium! 二.我的解答 乍一看,这个是计算最短路径的,迪杰斯特拉或者弗洛伊德算法都可以.不用这么复杂,同上一个题目一样: 刷题62. Unique Paths() 不多啰嗦,直接代码,注释中有原理: #include<iostream> #include<vector> using namespace std; class Solution{…

[BZOJ3817/UOJ42]Sum(类欧) 题面 BZOJ UOJ 题解 令\(x=\sqrt r\),那么要求的式子是\[\sum_{d=1}^n(-1)^{[dx]}\] 不难发现,对于每个\(d\)而言的取值只和\([dx]\)的奇偶性相关. 如果\(x\)是个整数,也就是\(r\)是完全平方数的时候,显然是可以直接算答案的. 计算答案的时候显然之和有几个奇数或者几个偶数相关(只要求一个另外一个就是补集) 比如说我们来求有几个是偶数,那么要满足的条件就是\([dx]=2*[\frac{…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=4676 Sum Of Gcd Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 908    Accepted Submission(s): 438 Problem Description Given you a sequence of numb…

点此看题面 大致题意: 多组询问,求\(\sum_{i=L}^R\sum_{j=i+1}^Rgcd(i,j)\). 推式子 这道题我们可以考虑,每个因数\(d\)被统计答案的次数,肯定与其出现次数有关. 设它出现次数为\(cnt_d\),则可以猜测答案为: \[\sum_{d=1}^nd\cdot C_{cnt_d}^2\] 这显然是错的,因为\(d\)虽作为公因数,却不一定是最大公约数. 所以就可以考虑容斥. 对于一个统计过的数\(x\),按照我们先前的做法,对于任意\(d|x\),\(d\)…

bzoj3944 题目描述 输入 一共T+1行 第1行为数据组数T(T<=10) 第2~T+1行每行一个非负整数N,代表一组询问 输出 一共T行,每行两个用空格分隔的数ans1,ans2 样例输入 6 1 2 8 13 30 2333 样例输出 1 1 2 0 22 -2 58 -3 278 -3 1655470 2 bzoj4805 同上,不需要求mu 题解 杜教筛 公式推导: 这里有一个难点(其实也不能算难),就是由枚举d|i到枚举j≤⌊n/i⌋.此时可以看作下面语句的i是上面语句的i/d,…

Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3980    Accepted Submission(s): 1620 Problem Description   Sample Input 2   Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Notice You can only move either down or right at any point in time! Dynamic programming is ultilized…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 这道题跟之前那道Dungeon Game 地牢游戏 没有什么太大的…

Problem: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Summary: 想要从m*n的整型数矩阵左上角…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 思路:由于只能向两个方向走,瞬间就没有了路线迂回的烦恼,题目的难度…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 做这题之前,建议先看一下Leetcode Triangle 这题看…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Summary: DP, calucate the minimun…

题目链接 Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 典型的动态规划问题. 设dp[i][j]表示从左上角到g…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. You can only move either down or right at any point in time. public class Solution { /** * @param gr…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 解题思路: dp问题,和上一题一样,JAVA实现如下: stati…