codeforces987E(想法?)

本文介绍了一种通过选择排序来确定特定序列被打乱方式的方法。通过计算最少交换次数并判断其奇偶性,可以推断出是Petr还是Um_nik进行了打乱操作。文章还提供了一个实现该算法的C++示例代码。

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题意:给出一个1到n的序列,Petr打乱了3n次,Um_nik打乱了7n+1次,现你有打乱后的数列,求是谁打乱的。
分析:比赛的时候怕是失了智在那边想随机算法…事实上用选择排序,找出最少交换次数,然后判断奇偶性就可以了……

    for(int i=1;i<=n;i++){
        cin>>now[i];//目前所占i这个位置的数字
        pos[now[i]]=i; //pos[i]记录i这个数字所在的位置
    }

AC代码:

#include<bits/stdc++.h>
using namespace std;
int now[1000000+7],pos[1000000+7];
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>now[i];//目前所占i这个位置的数字
        pos[now[i]]=i; //pos[i]记录i这个数字所在的位置
    }
    long long sum=0;
    for(int i=1;i<=n;i++){
        if(now[i]==i) continue;
        int temp1,temp2;
        temp1=now[i],temp2=pos[i];//temp1记录当前这个位置上的数字,temp2记录这个位置上原来应该有的数字
      //  cout<<"before swap: "<<now[i]<<" "<<now[pos[i]]<<endl;
        swap(now[i],now[pos[i]]);
      //  cout<<"after swap: "<<now[i]<<" "<<now[pos[i]]<<endl;
        pos[i]=i;
        pos[temp1]=temp2;
       // cout<<"new position for "<<temp1<<" is "<<temp2<<endl;
        sum++;
    }
    if((n*3-sum)&1) cout<<"Um_nik"<<endl;
    else cout<<"Petr"<<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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