Dropping testsadmin管理员组文章数量:1558103
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9338 | Accepted: 3272 |
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).
首先要注意这题不能用单位价值的排序来求答案是错的,然后二分求得时候要注意精度,终止条件是俩个均不为0
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int n, m;
const int N = 1100;
int a[N], b[N];
double v[N];
const int inf = 1000000010;
int judge(double k);
int main()
{
while(scanf("%d %d", &n, &m),n!=0||m!=0)
{
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(int i=0;i<n;i++)
{
scanf("%d",&b[i]);
}
m=n-m;
double l=0, r=inf, mid;
for(int i=0;i<100;i++)
{
mid=(l+r)/2;
if(judge(mid))
{
l=mid;
}
else
{
r=mid;
}
}
r=r*100;
printf("%.0f\n",r);
}
return 0;
}
int judge(double k)
{
for(int i=0;i<n;i++)
{
v[i]=1.000*a[i]-1.000*b[i]*k;
}
sort(v,v+n);
double sum=0.000;
for(int i=0;i<m;i++)
{
sum+=v[n-i-1];
}
if(sum>=0.0001) return 1;
else return 0;
}
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