Codeforces 492E Vanya and Field 规律题

本文探讨了一种算法,旨在解决在n*n矩阵中,利用给定向量从任意起点出发,以最高效的方式采集最多苹果的问题。通过将矩阵划分为n个等价的点集合,算法有效识别最优起点,实现最大化的苹果采集数量。

题目链接:点击打开链接

给定n*n的矩阵(0,0)->(n-1, n-1) m个苹果(下面m行给出苹果坐标)(dx, dy) 向量。

任选一个起点,用这个向量在矩阵里跑,问最多能采摘多少个苹果(坐标是%n, 即超过矩阵时 (x%n, y%n))

输出起点。

思路:

把向量所在的点集写出来会发现一个起点一定经过了n个点,即至多只有n种起点

所以把点分成n个组即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
const int N = 1000050;
const double eps = 1e-9;
typedef long long ll;
int a[N], b[N];
int n, m, dx, dy;
int main(){
    while(cin>>n>>m>>dx>>dy){
        memset(b, 0, sizeof b);
        int x = 0, y = 0;
        for(int i = 0; i < n; i++){
            a[x] = y;
            x = (x+dx)%n; y = (y+dy)%n;
        }
        for(int i = 1; i <= m; i++){
            rd(x); rd(y);
            y = (y-a[x]+n)%n;
            b[y]++;
        }
        int ans = 0;
        for(int i = 0; i < n; i++)
            if(b[i]>b[ans])
            ans = i;
        printf("%d %d\n", 0, ans);
    }
	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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