CodeForces 492E Vanya and Field

本文探讨了在n*n矩形中,从任一点开始以指定速度向量移动,直到回到已访问位置的情况下,如何通过求解扩展欧几里得方程来确定初始点,以达到经过最多点数的目标。详细步骤包括计算速度向量与矩阵的圈状运动,以及利用扩展欧几得算法找出符合条件的初始点。

题意:

n*n的矩形内有m(10^5)个点  从任一点开始以速度向量(dx,dy)移动  直到走到曾经走过的位置  问  从哪里开始  可以经过最多的点数  dx和dy均与n互素

思路:

很容易想到以dx,dy移动的话  走的一定是一个圈  不会是出现“蝌蚪形”然后循环  注意题意最后一句  而且每个x坐标一定只经过一次

那么对于m个点  我们可以求出它是(0,y)这个点可以到达的  求解方法就是用拓展欧几里德解下面方程

(x+dx*k)%mod=xi , (y+dy*k)%mod=yi 其中x=0 那么可以先解出k  把k带到二式解出y

最后统计对于每个(0,y)可以到达的点数就是答案

代码:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<bitset>
using namespace std;
typedef long long LL;
#define N 2000010

int n, m, ans;
LL dx, dy, posy;
map<LL, int> cnt;

long long extend_gcd(long long a, long long b, long long &x, long long &y) {
    if (a == 0 && b == 0)
        return -1;
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    long long d = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

int main() {
    cin >> n >> m >> dx >> dy;
    LL k, tmpk;
    extend_gcd(dx, n, k, tmpk);
    while (m--) {
        LL x, y;
        cin >> x >> y;
        if (x != 0) {
            tmpk = k * x % n * dy % n;
            if (tmpk <= y)
                y = y - tmpk;
            else
                y = y + n - tmpk;
        }
        y %= n;
        cnt[y]++;
        if (cnt[y] > ans) {
            ans = cnt[y];
            posy = y;
        }
    }
    cout << "0 " << posy << endl;
    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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