【luoguP2989】[USACO10MAR]对速度的需要Need For Speed

题目描述

最大化平均值

二分一个\(x\)

\(check\):

\(\frac{F+\sum_{i=1}^{n} X_{i} \times F_{i}}{M+\sum_{i=1}^{n} X_{i} \times M_{i}}\geq x\)

\(F+\sum_{i=1}^{n} X_{i} \times F_{i} \geq M \times x+\sum_{i=1}^{n} X_{i} \times M_{i} \times x\)

\(F-M \times x+\sum_{i=1}^{n} (X_{i} \times F_{i}-X_{i} \times M_{i} \times x)\ge 0\)

贪心的将\(1\)到\(n\)中大于\(0\)的\((X_{i} \times F_{i}-X_{i} \times M_{i} \times x)\)加起来，判断一下就可以了

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=10010;

int n,M,F,f[MAXN],m[MAXN];

inline bool check(double x){
    double sum=F-M*x;
    for(int i=1;i<=n;++i)
        sum+=max(0.0,f[i]-m[i]*x);
    return sum>=0;
}

int main()
{
    scanf("%d%d%d",&F,&M,&n);
    for(int i=1;i<=n;++i)
        scanf("%d%d",&f[i],&m[i]);
    double l=0,r=1e18;
    while(r-l>0.001){
        double mid=(l+r)/2.0;
        if(check(mid)) l=mid;
        else r=mid;
    }
    bool flag=0;
    for(int i=1;i<=n;++i)
        if(f[i]-m[i]*l>0) printf("%d\n",i),flag=1;
    if(!flag) puts("NONE");
    return 0;
}

原文地址：https://www.cnblogs.com/yjkhhh/p/11744150.html