HackerRank - powers-game-1 [博弈论] 题意 给出 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5 * 2^n 这一串东西 ,然后有两个玩家,*号是可以被替换掉的东西,可以换成+ 或者 - 然后最后的式子求出来后MOD 17 如果最后的结果 == 0 则 P2 wins 否则 P1 wins 思路 因为MOD 17 根据同余定理,我们可以 在替换*号之前就MOD 比如 2^1 2^2 2^3 2^4 2^5 2^6 2^7 2^8 2 4 8 16 15 1…

将整个游戏可以划分成若干个互不相交的子游戏. 每个子游戏的sg值只与其中的数的个数有关.而这个数不会超过30. 于是可以预处理出这个sg值表. 然后从1到n枚举,对<=sqrt(n)的部分,用个set判重. 对于大于sqrt(n)的部分,统计其中不包含在之前已经划分出来的子游戏内的数的个数,如果是奇数,就再异或上1. /* #include<cstdio> #include<cstring> #include<set> #include<map> us…

背景: 昨天看了<最强大脑>,由于节目比较有争议性,不知为什么,作为一名感性的人,就想试一下如果自己理性分析会是怎样的呢? 过程是这样的: 中国队(3人)VS英国队(4人). 1:李建东(队长)出战,[并说中国队不胜就再不参加最强大脑]3局过后,打平,双方都没脑力进行下一轮,所以评委各得1分,结果:1V1. 2:苏XX(忘名了)出战,打败对手,结果:2V1. 3:申一帆出战,失败,结果2V2平(同时申一帆情绪失控离开节目现场,经节目组一番说辞后回归节目) 问题来了:最后一战,谁出站,在大屏幕播…

Euclid's Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9033   Accepted: 3695 Description Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser o…

题目传送门:https://www.hackerrank.com/challenges/unique-colors 感谢hzq大神找来的这道题. 考虑点分治(毕竟是路经统计),对于每一个颜色,它的贡献是独立的.我们可以在一次点分治中合在一块处理(为什么时间复杂度是对的呢,因为我们每次改动只会根据当前点的颜色进行变动,而不是所有颜色对着它都来一遍).每次先对重心单独计算答案贡献,此时也将当前区域的各个答案贡献计算出来,并以此为基础(之后称之为基准贡献,即代码中的tot).对于每一棵子树,我们先df…

Game Theory Reveals the Future of Deep Learning Carlos E. Perez Deep Learning Patterns, Methodology and Strategy @ IntuitionMachine.com 译自:https://medium.com/intuitionmachine/game-theory-maps-the-future-of-deep-learning-21e193b0e33a#.2vjbrl5di 若你一直fo…

一些比较水的博弈论...(为什么都没有用到那什么SG呢....) TYVJ 1140  飘飘乎居士拯救MM 题解: 歌德巴赫猜想 #include <cmath> #include <cstdio> int n, a, b, ta, tb; inline bool isPrime(int x){ ) return true; ; i<=sqrt(x); i++) if (!(x%i)) return false; return true; } inline int getTi…

C. The Game Of Parity time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n cities in Westeros. The i-th city is inhabited by ai people. Daenerys and Stannis play the following game:…

Firt thought: an variation to LCS problem - but this one has many tricky detail. I learnt the solution from this link:https://github.com/wangyongliang/Algorithm/blob/master/hackerrank/Strings/Square%20Subsequences/main.cc And here is his code with my…

It is about how to choose btw. BFS and DFS. My init thought was to DFS - TLE\MLE. And its editorial gives a very neat BFS based idea which costs much less memory. https://www.hackerrank.com/challenges/beautiful-path/editorial…

It is marked as a NPC problem. However from the #1 code submission (https://www.hackerrank.com/CharlesOfria), it looks pretty much like a Brutal-Force (or simulation) based solution, mixed with some greedy strategies. To me the other NPC "Queens Revi…

传送门 今天在HackerRank上翻到一道高精度题,于是乎就写了个高精度的模板,说是模板其实就只有乘法而已. Extra long factorials Authored by vatsalchanana on Jun 16 2015 Problem Statement You are given an integer N. Print the factorial of this number. N!=N×(N−1)×(N−2)×⋯×3×2×1 Note: Factorials of N>20…

Ivan and Powers of Two Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 404C Description Valera had an undirected connected graph without self-loops and multiple edges consisting of n ver…

http://poj.org/problem?id=2234 博弈论真是博大精深orz 首先我们仔细分析很容易分析出来,当只有一堆的时候,先手必胜:两堆并且相同的时候,先手必败,反之必胜. 根据博弈论的知识(论文 张一飞:<由感性认识到理性认识——透析一类搏弈游戏的解答过程>) 局面可以分解,且结果可以合并. 局面均是先手 当子局面是 胜 和 败,那么局面则为胜 当子局面是 败 和 胜,那么局面则为胜 当子局面是 败 和 败,那么局面则为败 当子局面为 胜 和 胜,那么局面为不确定 而这些性质…

Great learning for me:https://www.hackerrank.com/rest/contests/master/challenges/lucky-numbers/hackers/turuthok/download_solution Basically it is memorized search application. And we follow a discrete strategy: split it into digits and go digit by di…

This is 'Difficult' - I worked out it within 45mins, and unlocked HackerRank Algorithm Level 80 yeah! So the idea is straight forward: 1. sort the input array and calculate partial_sum()2. find the negative\positive boundary with the accumulated give…

A sly knapsack problem in disguise! Thanks to https://github.com/bhajunsingh/programming-challanges/tree/master/hackerrank/algorithms/the-indian-jobLesson learnt: The Italian\Indian job is two-way 01 Knapsack. And some complext problem can be convert…

The most interesting, flexible and juicy binary tree problem I have ever seen. I learnt it from here: https://codepair.hackerrank.com/paper/5fIoGg74?b=eyJyb2xlIjoiY2FuZGlkYXRlIiwibmFtZSI6IkJsdWVCaXJkMjI0IiwiZW1haWwiOiJoZWFsdGh5dG9ueUBnbWFpbC5jb20ifQ%…

Something to learn: Rotation ops in AVL tree does not require recursion. https://github.com/andreimaximov/hacker-rank/blob/master/data-structures/tree/self-balancing-tree/main.cpp node *create_node(int val) { node *pNew = new node; pNew->val = val; p…

Powers of Two 题意: 让求ai+aj=2的x次幂的数有几对,且i < j. 题解: 首先要知道,排完序对答案是没有影响的,比如样例7 1一对,和1 7一对是样的,所以就可以排序之后二分找2的x次幂相减的那个数就好了,注意:打表时2的x次幂不能只小于1e9,因为有可能是2个5e8相加,之后就超出了1e9,但是你打表的时候又没有超出1e9的那个2的x次幂,所以答案总是会少几个.所以尽量就开ll 能多打表就多打. 代码: #include <bits/stdc++.h> usin…

文章原地址:http://blog.csdn.net/zhangxiang0125/article/details/6174639 博弈论:是二人或多人在平等的对局中各自利用对方的策略变换自己的对抗策略,达到取胜目标的理论.博弈论是研究互动决策的理论.博弈可以分析自己与对手的利弊关系,从而确立自己在博弈中的优势,因此有不少博弈理论,可以帮助对弈者分析局势,从而采取相应策略,最终达到取胜的目的. 博弈论分类:(摘自百度百科) (一)巴什博奕(Bash Game):只有一堆n个物品,两个人轮流从这堆…

思路:首先用Tarjan算法找出树中的环,环为奇数变为边,为偶数变为点. 之后用博弈论的知识:某点的SG值等于子节点+1后的异或和. 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<vector> #include<cstring> using namespace std; int ans; vector<…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1404 一看就是博弈论的题目,但并没有什么思路,看了题解,才明白 就是求六位数的SG函数,暴力一遍,打表就OK. 具体的操作是先找P态,即最终无法移动的状态,可知无数可取是P态,0是N态,1是P态,然后从1开始进行暴力, 所有可以到!sg[i]的点标记为N态,暴力过程为标记一步可以到sg[i]的数,包括两类: 一类是仅某一位数字不同,提取方法比较巧妙: ; --i){ int m = x; ; ; j…

https://www.hackerrank.com/contests/w9/challenges/lexicographic-steps 这题还是折腾很久的.题目意思相当于,比如有两个1两个0,那么找组成的数里第k大的.想法就是,如上例,假如K为4,那么先看后两位够了么C(2,2)=1,不够,那么看后三位C(3,2)=3,也不够,后四位是C(4,2)=6,够了,那么第一个1在倒数第4位.然后减去C(3,2)继续做. #include <iostream> #include <vecto…

https://www.hackerrank.com/contests/infinitum-aug14/challenges/jim-beam 学习了线段相交的判断法.首先是叉乘,叉乘的几何意义是有向的平行四边形的面积(除以2就是三角形的面积).如果ABD和ABC正负相反,说明C和D在AB两侧,同样的,再判断A和B是否在CD两侧即可.当某三角形面积为0时,需要判断是否在线段上. #include <iostream> using namespace std; typedef long long…

https://www.hackerrank.com/challenges/service-lane 用RMQ做的,其实暴力也能过~ #include <iostream> #include <vector> #include <cmath> using namespace std; int main() { int n, t; cin >> n >> t; vector<int> vec(n); for (int i = 0; i…

https://www.hackerrank.com/contests/w1/challenges/volleyball-match 此题不错,首先可以看出是DP,S(x, y)= S(x - 1, y) + S(x, y - 1).然后比赛结束状态需要认真判断.三来,最后数据量很大(接近10^9)远超一般DP的数据量,分配数组都不行,里面是有规律的.下面是大数据失败的代码: #include <vector> #include <iostream> #include <cs…

https://www.hackerrank.com/contests/w1/challenges/maximizing-xor/ 找了半天规律,答案竟然是暴力,伤感.我找到的方法是利用规律2^x XOR 2^x - 1会最大,感觉稍微效率高点. int maxXor(int l, int r) { if (l == r) return 0; int p = 1; for (int i = 0; i <= 10; i++) { if (p * 2 > r) break; p *= 2; } i…

https://www.hackerrank.com/contests/w2/challenges/cut-the-tree 树分成两部分,求两部分差最小.一开始想多了,后来其实求一下总和,求一下分部的和就行了. #include <cstdlib> #include <climits> #include <algorithm> #include <iostream> #include <vector> using namespace std;…

https://www.hackerrank.com/contests/w2/challenges/manasa-and-stones 简单题. #include<iostream> using namespace std; int main() { int T; cin >> T; while (T--) { int n, a, b; cin >> n >> a >> b; // a < b if ( a > b) { int tm…