POJ2976_Dropping tests_二分最大化平均值

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12121		Accepted: 4231

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

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).





有一点不同的是本题是要求去掉 c 个，所以那个循环是 i < n - c 。

#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

const int maxn = 1010;
int n, c;
int a[maxn], b[maxn];
double dis[maxn];// a - mid * b

bool C(double x)
{
for(int i= 0; i< n; i++)
dis[i] = a[i] - x * b[i];

sort(dis, dis+n);

double sum = 0;
for(int i= c; i< n; i++)
sum += dis[i];

//如果最大的 n-c 项和为正， 说明 x 合法
return sum >= 0;
}

int main ()
{
while(1){
scanf("%d %d", &n, &c);
if(n == 0 && c == 0) break;

for(int i= 0; i< n; i++)
scanf("%d", a+i);

for(int i= 0; i< n; i++)
scanf("%d", b+i);

double lb = 0.0, ub = 1.0;
for(int k= 0; k< 100; k++){
double mid = (lb + ub) / 2;

if(C(mid)) lb = mid;
else ub = mid;
}

//这里的处理方式要学习
printf("%.0f\n", 100 * lb);
}

return 0;
}