CodeForces 768E Game of Stones (Nim博弈)

本文解析了CodeForces平台上的768E题目的解题思路,该题涉及博弈论中的Nim游戏变化形式。通过计算每堆石子最多能不重复拿取的次数,并将其视为Nim博弈中的石子数进行异或操作,从而判断后手玩家是否能赢得比赛。

题目链接:http://codeforces.com/problemset/problem/768/E

题意:给定n堆石子,第i堆石子有a[i]个,两个玩家交替从n堆中取石子,不过如果一个玩家在第i堆中取过m个石子,则另一个玩家则不能在第i堆再取m个石子,最后不能取石子的人为输,jon赢则输出YES否则输出NO,jon为后手

思路:对于玩家来说,第i堆石子不能拿以前拿过的数量,求出第i堆石子最多不重复能够拿的次数,然后类比nim博弈异或求和即可。

          关于为什么求出最多不重复能够拿的次数就能够利用Nim博弈来求解:把第i堆最多不重复拿的次数类比成Nim博弈中第i堆石子的个数

#include <bits/stdc++.h>
using namespace std;

int n,a[1100000],b[1100000];

int main(){
    scanf("%d",&n);
    for(int i = 0; i < n; i ++) scanf("%d",&a[i]);
    for(int i = 0; i < n; i ++){
        int k = 1, cnt = 0;
        while(1){
            a[i] -= k; k ++;
            if(a[i] < 0) break;
            cnt ++;
        }
        b[i] = cnt;
    }
    int res = 0;
    for(int i = 0; i < n; i ++){
        res ^= b[i];
    }
    if(!res) printf("YES\n");
    else printf("NO\n");
    return 0;
}


### 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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