codeforces 768E Game of Stones

这是一个关于博弈论的游戏问题,玩家从n堆石头中按特定规则移除石头,每堆石头只能移除有限次。先手玩家(Sam)已经开始了游戏,后手玩家(Jon)想知道他是否有可能获胜。解决方案涉及到求解每堆石头的最大移除次数,并通过优化策略来确定胜利条件。如果Jon可以获胜,则输出'YES',否则输出'NO'。

E. Game of Stones
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Sam has been teaching Jon the Game of Stones to sharpen his mind and help him devise a strategy to fight the white walkers. The rules of this game are quite simple:

The game starts with n piles of stones indexed from 1 to n. The i-th pile contains si stones.
The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0 stones does not count as a move.
The player who is unable to make a move loses.
Now Jon believes that he is ready for battle, but Sam does not think so. To prove his argument, Sam suggested that they play a modified version of the game.

In this modified version, no move can be made more than once on a pile. For example, if 4 stones are removed from a pile, 4 stones cannot be removed from that pile again.

Sam sets up the

### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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