​​https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1257​​

很裸的分数规划 二分答案 得如下转换

sum(pi)/sum(wi)=ans? => sum(pi)=ans*sum(wi)? => sum(pi-ans*wi)=0?

发现ans具有二分性 如果ans太小则sum(pi-ans*wi)会小于0

二分出ans后按pi-ans*wi升序排序 取前k大的 要是凑不够0那就是太小 凑的够那就说明ans还有上升空间

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const double N=1e18;
const double eps=1e-6;
const int maxn=50010;

struct node
{
    ll w;
    ll p;
    double v;
};

node ary[maxn];
ll a,b,gcd;
int n,k;

bool cmp(node n1,node n2)
{
    return (double)(n1.p)-n1.v*(double)(n1.w)>(double)(n2.p)-n2.v*(double)(n2.w);
}

bool judge(double lim)
{
    double sum;
    int i;
    for(i=1;i<=n;i++) ary[i].v=lim;
    sort(ary+1,ary+n+1,cmp);
    sum=0.0;
    for(i=1;i<=k;i++) sum+=(double)(ary[i].p)-ary[i].v*(double)(ary[i].w);
    if(sum<0.0) return false;
    else
    {
        a=0,b=0;
        for(i=1;i<=k;i++) a+=ary[i].p;
        for(i=1;i<=k;i++) b+=ary[i].w;
        gcd=__gcd(a,b);
        a/=gcd,b/=gcd;
        return true;
    }
}

int main()
{
    double l,r,m;
    int i;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;i++) scanf("%lld%lld",&ary[i].w,&ary[i].p);
    l=0.0,r=N;
    while(r-l>eps)
    {
        m=(l+r)/2.0;
        if(judge(m)) l=m;
        else r=m;
    }
    printf("%lld/%lld\n",a,b);
    return 0;
}