poj 2976 Dropping tests（01分数规划+二分）

【题目大意】：给n个二元组，叫你从中去掉k个，使得sigema(100*a[i]-b[i])最大

【解题思路】：我们不妨令ans=sigema(100*a[i]-b[i])...得到sigema(a[i])*100-sigema(b[i])*ans=0;

   然后...接下就是二分...求max就可以了...

【代码】：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define linf 1LL<<60
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long

double a[1100],b[1100],tmp[1100];
int n,k;
double summ;

bool cmp(const double &a,const double &b){
    return a>b;
}

bool check(double mid){
    for (int i=0; i<n; i++) tmp[i]=a[i]*100-b[i]*mid;
    sort(tmp,tmp+n,cmp);
    double summ=0.0;
    for (int i=0; i<n-k; i++) summ+=tmp[i];
    if (summ>0) return true;
    else return false;     
}

double solve(){
    double low=0,high=100,mid;
    while (low+eps<high){
        mid=(low+high)/2.0;
        if (check(mid)) low=mid;
        else high=mid;
    }
    return mid;
}

int main() {
    while (~scanf("%d%d",&n,&k)){
        if (n==0 && k==0) break;
        for (int i=0; i<n; i++) scanf("%lf",&a[i]);
        for (int i=0; i<n; i++) scanf("%lf",&b[i]);
        printf("%.0f\n",solve());
          
    }
    return 0;
}