Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9329 | Accepted: 3271 |
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
将条件写为(n&&k)。无限WA。
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int MAXN=;
const double EPS=1.0e-6;
int n,k;
int a[MAXN],b[MAXN];
bool test(double x)
{
double y[MAXN];
for(int i=;i<n;i++)
{
y[i]=a[i]-x*b[i];
}
sort(y,y+n);
double sum=;
for(int i=k;i<n;i++)
{
sum+=y[i];
}
return sum>=0.0;
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
if(n==&&k==) break;
for(int i=;i<n;i++) scanf("%d",&a[i]);
for(int i=;i<n;i++) scanf("%d",&b[i]);
double l=;
double r=0x3f3f3f3f;
while(fabs(r-l)>EPS)
{
double mid=(l+r)/;
if(test(mid)) l=mid;
else r=mid;
}
int res=(int)(l*+0.5);
printf("%d\n",res);
}
return ;
}