POJ 2976 Dropping tests【分数规划】【二分搜索】

题目来戳呀

Description

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

100∗∑ni=1ai∑ni=1bi

.

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 n positive 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

3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

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).

题意

n门课程，给出每门的得分a及每门的满分b，求得分之和/满分之和的最大值。同时还给出k,可以去掉k门课程来保证更大的得分平均值。

想法

就是分数规划啊+_+
令r = ∑a[i] * x[i] / (b[i] * x[i]) 则必然∑a[i] * x[i] - ∑b[i] * x[i] * r= 0；（条件1）
并且任意的 ∑a[i] * x[i] - ∑b[i] * x[i] * max(r) <= 0 (条件2，只有当∑a[i] * x[i] / (b[i] * x[i]) = max(r) 条件2中等号才成立)
然后枚举r就可以了。
注意要排序，因为要拿走k 门（注意不是求k门。）

#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;double a[1100],b[1100],f[1100];;const double acc=1e-7;int main(){    int n,k;    double mid,ans;    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]);        double l=0.0,r=1.0;        while(r-l>acc)        {            mid=(l+r)*1.0/2;            for(int i=0; i<n; i++)            {                f[i]=a[i]-mid*b[i];            }            sort(f,f+n);            ans=0;            for(int j=k; j<n; ++j)            {///逃课k门，成绩是n-k门                ans+=f[j];            }            if(ans>0)                l=mid;            else                r=mid;        }        printf("%.f\n",mid*100);///mid才是我们要的答案 不是ans+_+    }    return 0;}

ps：算了算了 身体好了再补吧 头疼的要炸裂QAQ