Codeforces 725D[Contest Balloons]【贪心】

本文介绍了一种竞赛编程策略,通过给予他人资源以优化自己排名的方法。使用小根堆维护潜在上浮团队,确保分配资源后的最优排名。

Description

One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn’t matter and teams are sorted only by the number of balloons they have. It means that one’s place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.

You should know that it’s important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn’t considered in the standings.

A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It’s guaranteed that ti*doesn’t exceed *wi so nobody floats initially.

Limak is a member of the first team. He doesn’t like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can’t give away more balloons than his team initially has.

What is the best place Limak can get?

题解

如果我要把气球分给别人,那么我一定是优先分给那些排在我前面的,并且需要最少的气球就可以上天的,并且,我每次分掉一些气球后会有另一些人排到我的前面去,所以,只要用一个小根堆维护一下,从每次的答案中刷出最小值就是答案。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 300006
#define LL long long
using namespace std;
inline char nc(){
    static char buf[100000],*i=buf,*j=buf;
    return i==j&&(j=(i=buf)+fread(buf,1,100000,stdin),i==j)?EOF:*i++;
}
inline LL _read(){
    char ch=nc();LL sum=0;
    while(!(ch>='0'&&ch<='9'))ch=nc();
    while(ch>='0'&&ch<='9')sum=sum*10+ch-48,ch=nc();
    return sum;
}
struct data{
    LL t,w;int id;
    bool operator >(const data&b)const{return w-t>b.w-b.t;};
}a[maxn];
int n,ans;
priority_queue<data,vector<data>,greater<data> >heap;
bool cmp(data x,data y){
    return x.t>y.t||(x.t==y.t&&x.id<y.id);
}
int main(){
    freopen("balloons.in","r",stdin);
    freopen("balloons.out","w",stdout);
    n=_read();
    for(int i=1;i<=n;i++)a[i].t=_read(),a[i].w=_read()+1,a[i].id=i;
    sort(a+1,a+1+n,cmp);
    int p;
    for(int i=1;i<=n;i++) if(a[i].id==1){
        p=i;break;
    }
    for(int i=1;i<p;i++)heap.push(a[i]);
    ans=p-1;
    int num=p-1,j=p+1;
    while(!heap.empty()&&a[p].t){
        data c=heap.top();
        if(a[p].t>=c.w-c.t){
            heap.pop();num--;
            a[p].t-=c.w-c.t;
            while(j<=n&&a[j].t>a[p].t){
                heap.push(a[j++]);
                num++;
            }
            ans=min(ans,num);
        }else break;
    }
    printf("%d",ans+1);
    return 0;
}
### 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?
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