CodeForces 385E

本文介绍了一个趣味编程问题:通过矩阵快速幂的方法计算一只熊在n*n森林中随时间变化的位置。熊的位置受特定规则控制,包括速度与位置的变化及森林中growingraspberry高度的增长。

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题意:
有一个熊在n*n的森林里,初始位置在(sx,sy),每个位置都有growing raspberry
每个growing raspberry初始高度是x+y
然后每秒会先发生速度变化,假设当前位置是(x,y)
t=growing raspberry的高度
速度会从(vx,vy)->(vx+t,vy+t)
然后发生位置变化(x,y)->((x+vx-1)%MOD+1,(y+vy-1)%MOD+1)
然后growing raspberry高度++
问你t秒后熊的位置
思路:
先把x,y变成0~n-1便于取模
然后就是普通的矩阵快速幂
(x0,y0,t0,Vx0,Vy0,1)->
(X0+Vx0+X0+Y0+t0+2,y0+Vx0+X0+Y0+t0+2,t0+1,Vx0+X0+Y0+t0+2,Vy0+X0+Y0+t0+2,1)

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x&(-x))
typedef long long LL;
const int maxn = 100005;
const int inf=(1<<28)-1;
#define Matrix_Size 10
LL MOD;
int Size;
struct Matrix
{
    LL mat[Matrix_Size][Matrix_Size];
    void clear()
    {
        memset(mat,0,sizeof(mat));
    }
    void output()
    {
        for(int i = 0;i < Size;i++)
        {
            for(int j = 0;j < Size;j++)
                printf("%d ",mat[i][j]);
            printf("\n");
        }
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret;
        for(int i = 0;i < Size;i++)
            for(int j = 0;j < Size;j++)
            {
                ret.mat[i][j] = 0;
                for(int k = 0;k < Size;k++)
                {
                    long long tmp = (long long)mat[i][k]*b.mat[k][j]%MOD;
                    ret.mat[i][j] = (ret.mat[i][j]+tmp);
                    if(ret.mat[i][j]>=MOD)
                        ret.mat[i][j] -= MOD;
                    if(ret.mat[i][j]<0)//注意是否需要MOD 
                        ret.mat[i][j] += MOD;
                }
            }
        return ret;
    }
};
Matrix pow_M(Matrix a,long long n)
{
    Matrix ret;
    ret.clear();
    for(int i = 0;i < Size;i++)
        ret.mat[i][i] = 1;
    Matrix tmp = a;
    while(n)
    {
        if(n&1)ret = ret*tmp;
        tmp = tmp*tmp;
        n>>=1;
    }
    return ret;
}
int Tmp[36]={
2,1,1,1,0,2,
1,2,1,0,1,2,
0,0,1,0,0,1,
1,1,1,1,0,2,
1,1,1,0,1,2,
0,0,0,0,0,1};
int main()
{
    LL n,sx,sy,dx,dy,t;
    scanf("%lld%lld%lld%lld%lld%lld",&n,&sx,&sy,&dx,&dy,&t);
    MOD=n;
    if(t==0)
    {
        printf("%lld %lld\n",sx,sy);
        return 0;
    }
    sx--,sy--;
    Size=6;
    Matrix A,B;
    A.clear();B.clear();
    for(int i=0;i<6;++i)
     for(int j=0;j<6;++j)
     A.mat[i][j]=Tmp[i*6+j];
    //A.output();printf("\n");
    B.mat[0][0]=sx;B.mat[1][0]=sy;
    B.mat[2][0]=0;B.mat[3][0]=dx;
    B.mat[4][0]=dy;B.mat[5][0]=1;
    A=pow_M(A,t);
    A=A*B;
    printf("%lld %lld\n",A.mat[0][0]+1,A.mat[1][0]+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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