codeforces 573E

本文介绍了一种通过贪心策略实现的最大序列求和算法。该算法通过对数组元素进行选择,确保所选序列的加权和最大。算法核心在于判断是否选择当前元素,以保证后续元素的权值增加能够弥补因选择当前元素而减少的值。

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题目描述

给你n个数,让你取出其中的某些数,使得剩下的数组成一个序列(不能调换顺序),序列的ki=1isi最大。

解题思路

多次贪心取最优值。

假设当前选的序列的前缀和为sumi
1.如果一个数aj在上一个序列没有选,但是ajnumj+sumnsumj0,numj表示aj排在序列的第几位。那么这样的话选这个数一定更优,因为如果选了aj,那么从j+1n选的数的权值都会增加,总共增加sumnsumj,如果这个增加量比选aj的失去量要大,那么,选,一定更优。

2.如果一个数aj在上一个序列有选,但是ajnumj+sumnsumj<0,一定不选更优。

参考代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define maxn 100005
#define ll long long
using namespace std;

ll a[maxn],s[maxn],ans,n;

bool cho[maxn];

int read(){
    int ret=0,ff=1;
    char ch=getchar();
    while (ch<'0' || ch>'9') {
        if (ch=='-') ff=-1;
        ch=getchar();
    }
    while (ch>='0' && ch<='9') {
        ret=ret*10+ch-'0';
        ch=getchar();
    }
    return ret*ff;
}

int main(){
    n=read();
    fo(i,1,n) {
        a[i]=read();
        cho[i]=1;
        s[i]=s[i-1]+a[i];
    }
    for(;;){
        int tot=1;
        bool bz=0;
        ll now=0;
        fo(i,1,n) {
            if (cho[i] && a[i]*tot+s[n]-s[i]<0) {
                cho[i]=0;
                bz=1;
            }
            else if (!cho[i] && a[i]*tot+s[n]-s[i]>=0) {
                cho[i]=1;
                bz=1;
            }
            if (cho[i]) {
                now+=a[i]*tot;
                tot++;
                s[i]=s[i-1]+a[i];
            }
            else s[i]=s[i-1];
        }
        ans=now;
        if (!bz) break;
    }
    cout<<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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