POJ 2976 Dropping tests(最大化平均值)

题意：给出n门课程，并给出每门课的所考的成绩和这门课的总成绩，可以去掉k门课， 求最大的平均分。

解题思路：平均分是sum(a[i]) / sum(b[i])，可行性函数判断的是， sum(a[i]) / sum(b[i]) >= x, 数学变换得sum(a[i] - x * b[i]) >= 0，所以每次进行可行性判断时先对a[i] - x * b[i]进行排序。

Description

In a certain course, you take n tests. If you get a_i out of b_i 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 a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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).

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 1005const int INF = 0x3f3f3f3f;int a[MAX_N];int b[MAX_N];double c[MAX_N];int n, k;bool f(double x){//    cout<<1<<endl;    for(int i = 0 ; i < n ; i++)    {        c[i] = a[i] - x * b[i];    }    sort(c, c + n);//    for(int i = 0 ; i < n ; i++)//    {//        cout<<c[i]<<" ";//    }//    cout<<endl;    double sum = 0;    for(int i = 0 ; i < n - k ; i++)    {        sum += c[n - i - 1];//        cout<<sum<<endl;    }//    cout<<sum<<endl;    return sum >= 0;}int main(){//    freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);    while(scanf("%d%d", &n, &k) && (n || k))    {        for(int i = 0 ; i < n ; i++)        {            a[i] = read();        }        for(int i = 0 ; i < n ; i++)        {            b[i] = read();        }        double lb = 0, ub = 1;        while(ub - lb > 1e-8)        {            double mid = (lb + ub) / 2;            if(f(mid)) lb = mid;            else ub = mid;        }        int ans = floor(ub * 100 + 0.5);        printf("%d\n", ans);    }    return 0;}