Codeforces 492E (math)

本文介绍了一种算法,用于计算在正方形区域内沿着特定向量行走时,经过最多特殊点的路径及其位置。通过分类计数的方法,实现了一个C++程序来解决这一问题。

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很有意思的一道题:
在正方形中按一个固定向量走,路过最多的特殊点。
注意,dx与n互质,dy与n互质,意味着走n步必然回到起点,且路径上同一个横坐标只会对应一个纵坐标。
所以,就简化成一个分类计数问题。

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

#define maxn 1000010

int ans[maxn];
int res[maxn];
int memory[maxn][2];

int main()
{
    int n,m,dx,dy;
    scanf("%d%d%d%d",&n,&m,&dx,&dy);
    memset(ans,0,sizeof(ans));
    memset(res,0,sizeof(res));

    int x = 0;
    int y = 0;

    for(int i=0;i<n;i++)
    {
        x = (x+dx)%n;
        y = (y+dy)%n;
        ans[x] = y;
    }

    for(int i=0;i<m;i++)
    {
        int tx,ty;
        scanf("%d%d",&tx,&ty);
        int loc = 0;
        if(ty < ans[tx]) loc = n - ans[tx] + ty ;
        else loc = ty - ans[tx];
        res[loc]++;
        memory[loc][0] = tx;
        memory[loc][1] = ty;
    }
    int maxans = 0;
    int maxloc = 0;
    for(int i=0;i<n;i++)
    {
        if(maxans < res[i])
        {
            maxans = res[i];
            maxloc = i;
        }
    }
    printf("%d %d\n",memory[maxloc][0],memory[maxloc][1]);
    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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