POJ 2976 二分之最大化平均值问题

In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating ai for all i. The third line contains npositive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

Sample Input

Sample Output

83
100

Hint

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

题目大意：从n对数选择n-k对数，使其运算上面的式子，然后求最大的平均值

分析：如果这题知道是最大化平均值的话，肯定就知道是通过二分来实现的，网上也有解释为什么这样做的，但是却很少有证明的，但是我无意中找到一篇大佬的详解，最大化平均值问题可以叫做01分数规划问题；复杂度为O（nlogn）

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
const int N = 3e5 + 10;
const int MOD = 0x3f3f3f3f;
int k,n;
bool vis[N];
double ans,temp;
struct node{
    double a,b;
}num[N];
double y[N];
bool solve(double x)
{
    rep(i,1,n)
    y[i]=num[i].a-x*num[i].b;
    sort(y+1,y+1+n);
    double sum=0;
    for(int i=n;i>k;i--) sum+=y[i];
    return sum>=0;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    while(scanf("%d%d",&n,&k)!=EOF,n+k){
            rep(i,1,n) scanf("%lf",&num[i].a);
            rep(i,1,n) scanf("%lf",&num[i].b);
            double l=0,r=MOD*1.0;
            for(int i=1;i<=100;i++){
                double mid=(l+r)/2;
                if(solve(mid)) l=mid;
                else r=mid;
            }
            printf("%.f\n",l*100.0);
    }
    return 0;
}

POJ 2976 二分之最大化平均值问题

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