POJ 2976-Dropping tests（01分数规划入门）

https://blog.csdn.net/Hhaile/article/details/8883652

这篇写得挺不错

这个就是入门的01分数规划，二分答案x，将ai-x*bi从大到小排序，去前n-k个，判断是否满足sigm(ai)-sigm(bi*x)>=0，满足 说明还有最优解即更新l=mid，否则更新r=mid;

#include <stdio.h>
#include <string.h>
#include<iostream>
#include <math.h>
#include <algorithm>
#include<functional>
using namespace std;
typedef long long ll;
#define eps 1e-6
const int N = 1e6 + 10;
int  n,k;
int a[N],b[N];
double c[N];
bool check(double  t)
{
    double sum=0;
    for (int i = 1; i <= n; i++)
        c[i] = a[i] - t*b[i];
    sort(c + 1, c + n + 1, greater<double>() );
    for (int i = 1; i <= n - k; i++)
        sum += c[i];
    return sum >= 0;
}
int main()
{
    while (cin >> n >> k && (n || k))
    {
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++)
            scanf("%d", &b[i]);
        double l = 0, r = 1;
        while (r - l > eps)
        {
            double mid = (l + r) / 2;
            if (check(mid)) l = mid;
            else r = mid;
        }
        printf("%.0lf\n", l*100);
    }
    return 0;
}