[二分]poj2976 Dropping tests

E - Dropping tests
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status
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

.

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

输入数组a和数组b，总共n对中可以去掉k对，求其中n-k对∑a[i]/∑b[i]的最大值。

假设我们要求这个式子的最大值，那么如何解这个式子呢？   假设这个式子的最大值，也就是最优解为val，即val=max（a*x/b*x）。 设置这样一个子问题：q(L)=a*x-b*x*L，然后我们分配向量x，来使q(L)取得最大值max，注意：这个式子不是一下能算出来的，因为a，b，x都是向量，如果向量是m维的，那么我们应该把算出来的m个数求和。这时q(L)的最大值max就分两种情况来看：①max<0 ，即a*x-b*x*L<0 ----> a*x<b*x*L  ----> (a*x)/(b*x) <L ， 那么无论怎么分配x都没法到达L值，即val<L,所以解要比L小，我们应该适当减小L的值再试。②max>=0 即val>=L,  且进一步还知道，当取到L=val时，有a*x-b*x*L=0，所以此时解要>=L，那么我们应当增大L的值再试。上面两种情况已经可以看出，我们可以二分L。说到这，上面说的要想策略，策略就是：分配x，使q(L)=a*x-b*x*L最大，我们需要求出n个t[i]=a[i]-b[i]*L，然后对t排序，取前n-k组最大的即可。

#include <iostream>#include <algorithm>#include<cstdio>using namespace std;const int MAXN = 100010;const int INF =0X3f3f3f3f;const double eps = 1e-6;int n,k;double a[MAXN],b[MAXN],t[MAXN];int cmp(double x,double y){    return x>y;}bool does(double num){    for(int i = 0; i< n;i++)        t[i] = a[i]- num * b[i];    sort(t,t+n,cmp);    double sum = 0.0;    for(int i = 0;i<n-k;i++)        sum += t[i];    return sum > 0;}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]);        double lb = 0.0,ub = 1.0;        while( ub-lb > eps )        {            double mid = (ub+lb)/2.0;            if (does(mid)) lb = mid;            else ub = mid;        }        printf("%d\n",(int)(ub*100+0.5));    }}