wyh的物品（01分数规划）

思路：有题可知最后的答案ans满足

$ans= \sum_{i=1}^mv_i /\sum_{i=1}^mw_i$ans=i=1∑m​vi​/i=1∑m​wi​
上面变形为
$\sum_{i=1}^mv_i-ans*\sum_{i=1}^mw_i>=0$i=1∑m​vi​−ans∗i=1∑m​wi​>=0

由上可知我们可以二分答案ans检测其正确性

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
#define ZY set<node>::iterator
#define lson root<<1
#define rson root<<1|1
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int imax_n = 1e5 + 7;
typedef pair<int,int> PII;
const int inf=0x3f3f3f3f;
//const int mod=1e9+7;
const int N=5e6+10;
const double esp=1e-10;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}

double a[N],b[N],c[N];
int main()
{
   int t;
   int n,k;
   read(t);
   while(t--)
   {
       read(n),read(k);
       for(int i=1;i<=n;i++)
       {
           read(a[i]),read(b[i]);
       }
       double l=0,r=10000000.0;
       while(r-l>esp)
       {
           double mid=(l+r)/2.0;
           for(int i=1;i<=n;i++)
            c[i]=1.0*b[i]-1.0*mid*a[i];
           sort(c+1,c+1+n,greater<double>());
           double sum=0.0;
           for(int i=1;i<=k;i++)
           {
               sum+=1.0*c[i];
           }
           if(sum>=0.0)l=mid;
           else r=mid;

       }
       printf("%.2f\n",l);


   }




    return 0;
}