codeforces 798d

本文针对 CodeForces 798D 题目,详细解析了一种有效的贪心策略。通过将两个数组合并并按特定规则排序,文章介绍了一种能够确保所选元素总和超过数组总和的方法。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目链接:http://codeforces.com/contest/798/problem/D
题目大意:
给两个数组,挑选n / 2 + 1个下标,使得a数组对应下标的和的两倍大于a数组的数之和,b数组对应下标的和的两倍大于b数组的数之和。

拿到题目想不到如何贪心才是最优的。。于是偷看了大神题解。

将两个数组拼在一起排序,按照a数组从大到小排序,先选择第一个,然后后面开始每两个,每次选择b数组中大的那个,如果最后是偶数的话,再选择最后一个。

考虑这样贪心为何是最优的:首先,b数组里选择了第一个,后面的每一个都是选择比较大的,所以肯定所选下标的总和是大于数组总和的。其次a数组里第一次开始选择的是第一个,假设后面每次选的都是后一个,也能满足从下标0开始,每次递增2个位置,每次i比i+1要大。所以该贪心是成立的orz。感觉很久不会贪心了。

附上代码:

/*
@resources: codeforces 798d
@date: 2017-4-27
@author: QuanQqqqq
@algorithm: greedy
*/

#include <bits/stdc++.h>

#define MAXN 100005
#define INF 1e10

using namespace std;

struct node{
    int a,b,idx;    
}num[MAXN];

int cmp(node a,node b){
    return a.a > b.a;
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i = 0;i < n;i++){
        scanf("%d",&num[i].a);
    }
    for(int i = 0;i < n;i++){
        scanf("%d",&num[i].b);
        num[i].idx = i + 1;
    }
    sort(num,num + n,cmp);
    printf("%d\n",n / 2 + 1);
    printf("%d",num[0].idx);
    for(int i = 1;i + 1 < n;i += 2){
        if(num[i].b > num[i + 1].b){
            printf(" %d",num[i].idx);
        } else {
            printf(" %d",num[i + 1].idx);
        }
    }
    if(n % 2 == 0){
        printf(" %d\n",num[n - 1].idx);
    }
}
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值