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 aifor 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).
题意:有N个考试,每个考试有ai和bi两个值,最后成绩由上面的公式求得。幸运的是,可以放弃K个科目,求最大化最后的成绩。
思路:这是一道简单的最大化平均值模板题,化简出(ai-mid*bi)>0,求n-k前项。
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #define INF 0x3f3f3f3f #include<algorithm> using namespace std; bool cmp(double x,double y) { return x>y; } int n,k; double y[1100]; int a[1100]; int b[1100]; bool C(double mid) { for(int i=0 ; i<n ; i++) { y[i]=a[i]-mid*b[i]; } sort(y,y+n,cmp); double sum=0; for(int i=0 ; i<n-k ; i++) { sum+=y[i]; } if(sum>=0) return true; return false; } int main() { while(scanf("%d%d",&n,&k)!=EOF) { if(n==0&&k==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 st=0,en=INF; for(int i=1 ; i<=100 ; i++) { double mid=(st+en)/2; if(C(mid)) st=mid; else en=mid; } printf("%.0f ",st*100); } }