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 0Sample Output
83 100Hint
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).
Source
Stanford Local 2005
题意很明了，给你一个n一个k
然后输入两排数，假设分别是a1,a2,a3......an以及b1,b2,b3.....bn
那么假设r=sigma(ai)/sigma(bi)现在我们可以扔掉k对（ai,bi），让r的值最大
比如当输入是
3 1 5 0 2 5 1 6 时我们可以扔掉（2，,6）
从而使r=(5+0)/(5+1)=83.33333%
四舍五入取整得到83
思路：
我们可以首先假设那个最大的r为一个定值
那么得到以下等式：100*sigma(a[i])/sigma(b[i])=r
变化一下：100*sigma(a[i])-r*sigma(b[i])=0
------------------->sigma(100*a[i]-r*b[i])=0
非常典型的单调函数！！！可以利用二分查找进行求解（很像是01划分树的问题）
可以先用100*a[i]-r*b[i]进行排序，然后选前面的n-k个数，然后再去求和，如果sum>0则l=mid
反之则r=mid（循环的次数也必须>=15，我先开始写的10，由于精度不够高所以wa了一次）
我的代码：
#include<stdio.h> #include<algorithm> using namespace std; struct node { double a; double b; double x; }; node in[1005]; bool cmp(node p,node t) { return p.x>t.x; } int main() { int n,k,i; double l,r,sum,mid; int time; while(scanf("%d%d",&n,&k)!=EOF) { if(n==0&&k==0) break; time=15; for(i=0;i<n;i++) scanf("%lf",&in[i].a); for(i=0;i<n;i++) scanf("%lf",&in[i].b); l=0; r=100; while(time--) { sum=0; mid=(l+r)/2; for(i=0;i<n;i++) in[i].x=100*in[i].a-mid*in[i].b; sort(in,in+n,cmp); for(i=0;i<n-k;i++) sum=sum+(100*in[i].a-mid*in[i].b); if(sum>0) l=mid; else r=mid; } printf("%.0lf/n",mid); } return 0; }
没有天分，只有勤奋！！
