牛客多校 gpa (二分)

最新推荐文章于 2025-02-07 22:27:48 发布

原创最新推荐文章于 2025-02-07 22:27:48 发布 · 286 阅读

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给出每门课的绩点和学分,最多删除k个课,问你最多均绩多少
一开始想着,就是贪心一下,某种排序,但是显然不可以(事实上在这上面撒比了很久甚至最后还在想)
想到了二分,但是不知道怎么二分,为什么呢,因为不知道二分的单调性在哪里,就不行了.
这题的核心就在于,不是排序后二分,而是二分每个之后排序,这样就可以单调确定每个点的价值.
二分是很灵活的东西,可以在任何地方出现.

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <iomanip>
#define debug(x) //std::cerr << #x << " = " << (x) << std::endl
using namespace std;
typedef long long LL;
const int MAXN = 2e5 + 17;
const int MOD = 1e9 + 7;
vector<pair<double,double > > all;
int c[MAXN],s[MAXN];
double a[MAXN];
int n,k;
bool judge(double x)
{
    for (int i = 0; i < n; ++i)
    {
        a[i] = 1.0*all[i].second*(1.0*all[i].first-x);
    }
    sort(a, a+n);
    double sum = 0;
    for (int i = k; i < n; ++i)
    {
        sum += a[i];
    }
    return sum>=0;
}
int main()
{
#ifdef noob
    freopen("Input.txt", "r", stdin);
    freopen("Output.txt", "w", stdout);
#endif
    cin>>n>>k;
    for (int i = 0; i < n; ++i)
    {
        scanf("%d",c+i);
    }
    for (int i = 0; i < n; ++i)
    {
        scanf("%d",s+i);
    }
    for (int i = 0; i < n; ++i)
    {
        all.push_back({s[i],c[i]});
    }
    double l = 0.0,r = 100000000+1;
    int tms = 100;
    for (int i = 0; i < tms; ++i)
    {
        double mid = (l+r)/2.0;
        if(judge(mid))
            l = mid;
        else
            r = mid;
    }
    debug(l);
    cout<<fixed<<setprecision(11)<<l<<endl;
    return 0;
}