【Codeforces249D】Donkey and Stars

本文介绍了一种基于输入点集的二维几何变换算法,通过将坐标系内的点转换到特定直线的一侧来实现。该算法首先读取一系列点,并对这些点进行线性变换,之后对变换后的点进行排序并使用单调栈进行处理,最终输出结果。

转换以后就是一个lis。
(程序非常简单易懂)

#include <bits/stdc++.h>
#define gc getchar()
#define ll long long
#define mid (l+r>>1)
#define N 100009
using namespace std;
int n,a,b,c,d,m,sta[N],top;
struct point
{
    ll x,y;
    bool operator <(const point &rhs) const
    {
        return (x<rhs.x)||(x==rhs.x)&&(y>rhs.y);
    }
}p[N];
int read()
{
    int x=1;
    char ch;
    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;
    int s=ch-'0';
    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';
    return s*x;
}
int main()
{
    n=read();
    a=read(),b=read(),c=read(),d=read();
    for (int i=1;i<=n;i++)
    {
        int x=read(),y=read();
        m++;
        p[m].x=(ll)y*b-(ll)x*a;
        p[m].y=(ll)x*c-(ll)y*d;
        if (p[m].x<=0||p[m].y<=0) m--;
    }
    sort(p+1,p+m+1);
    for (int i=1;i<=m;i++)
    {
        if (!top||p[i].y>p[sta[top]].y) sta[++top]=i;
        else
        {
            int l=1,r=top,ret=0;
            while (l<=r)
            {
                if (p[i].y<=p[sta[mid]].y) ret=mid,r=mid-1;
                else l=mid+1;
            }
            if (p[i].y<p[sta[ret]].y) sta[ret]=i;
        }
    }
    printf("%d\n",top);
    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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