标签：been plm test stream process sort osi 没有 问题
题目链接：http://poj.org/problem?id=2976
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20830 | Accepted: 7052 |
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 npositive 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).
题目大意：输入N K 下面有两行 每行有N个数 第一行代表N个a[i] 第二行代表N个b[i] 问你按照上面的公式算 去掉K个 得到的最大结果是多少
思路：自己并没有想到怎么二分 ，自己想到的是怎么贪心，但是感觉不行，就没去尝试了，其实想的到二分答案，但是怎么判断答案是否成立就没有想到了，这里用到的一种思想，分析表达式
以前我都是看到表达式就看一下的，从来没有给它仔细化简分析啥的 这一题算是一个教训吧！ 那么下面看一下这个表达式到底怎么回事
令r = ∑a[i] * x[i] / (b[i] * x[i]) 则必然∑a[i] * x[i] - ∑b[i] * x[i] * r= 0；（条件1）并且任意的 ∑a[i] * x[i] - ∑b[i] * x[i] * max(r) <= 0 (条件2，只有当∑a[i] * x[i] / (b[i] * x[i]) = max(r) 条件2中等号才成立)然后就可以枚举r , 对枚举的r， 求Q(r) = ∑a[i] * x[i] - ∑b[i] * x[i] * r 的最大值， 为什么要求最大值呢？ 因为我们之前知道了条件2，所以当我们枚举到r为max(r)的值时，显然对于所有的情况Q(r)都会小于等于0，并且Q(r)的最大值一定是0.而我们求最大值的目的就是寻找Q(r)=0的可能性，这样就满足了条件1，最后就是枚举使得Q(r)恰好等于0时就找到了max(r)。而如果能Q(r)>0 说明该r值是偏小的，并且可能存在Q(r)=0,而Q(r)<0的话，很明显是r值偏大的，因为max(r)都是使Q(r)最大值为0，说明不可能存在Q(r)=0了。
注意：二分真的很气人啊，这题用了以前经常用的二分板子，竟然错了。。。 真的不明白为啥，就是二分最后的结果是l 还是r 应该是这题有点问题，下面wa的代码也给出来吧
wa的：
#include<iostream> #include<algorithm> #include<stdio.h> using namespace std; typedef long long ll; const int maxn=1000+5; const double inf=1e15; const double eps=1e-7; int N,K; ll a[maxn],b[maxn]; double c[maxn]; bool cmp(const double x,const double y) { return x>y; } bool judge(double mid) { for(int i=0;i<N;i++)//找到最优的解 { c[i]=a[i]-mid*b[i]; } sort(c,c+N,cmp); double sum=0; for(int i=0;i<N-K;i++) sum+=c[i]; return sum>=0; } int main() { while(cin>>N>>K) { if(N==0&&K==0) break; for(int i=0;i<N;i++) cin>>a[i]; for(int i=0;i<N;i++) cin>>b[i]; double l=0,r=inf; double ans; while(r-l>=eps) { double mid=(r+l)/2; //cout<<mid<<endl; if(judge(mid)) { ans=mid; l=mid+eps; } else r=mid; //cout<<"ans: "<<ans<<endl; //cout<<"l: "<<l<<endl; //cout<<"r: "<<r<<endl; } printf("%.0lf\n", ans * 100); /* double mid; while(r-l>=eps) { mid=(l+r)/2; if(judge(mid)) l=mid; else r=mid; cout<<"l: "<<l<<endl; cout<<"r: "<<r<<endl; } printf("%.0lf\n",r*100); //printf("%.0lf\n", ans * 100); //cout<<(int)(ans*100+0.5)<<endl; */ } return 0; } /* */
ac的：
#include<iostream> #include<algorithm> #include<stdio.h> using namespace std; typedef long long ll; const int maxn=1000+5; const double inf=1e15; const double eps=1e-7; int N,K; ll a[maxn],b[maxn]; double c[maxn]; bool cmp(const double x,const double y) { return x>y; } bool judge(double mid) { for(int i=0;i<N;i++)//找到最优的解 { c[i]=a[i]-mid*b[i]; } sort(c,c+N,cmp); double sum=0; for(int i=0;i<N-K;i++) sum+=c[i]; return sum>=0; } int main() { while(cin>>N>>K) { if(N==0&&K==0) break; for(int i=0;i<N;i++) cin>>a[i]; for(int i=0;i<N;i++) cin>>b[i]; double l=0,r=inf; double ans; while(r-l>=eps) { double mid=(r+l)/2; //cout<<mid<<endl; if(judge(mid)) { ans=mid; l=mid+eps; } else r=mid; //cout<<"ans: "<<ans<<endl; //cout<<"l: "<<l<<endl; //cout<<"r: "<<r<<endl; } printf("%.0lf\n", r * 100); /* double mid; while(r-l>=eps) { mid=(l+r)/2; if(judge(mid)) l=mid; else r=mid; cout<<"l: "<<l<<endl; cout<<"r: "<<r<<endl; } printf("%.0lf\n",r*100); //printf("%.0lf\n", ans * 100); //cout<<(int)(ans*100+0.5)<<endl; */ } return 0; } /* */
标签：been plm test stream process sort osi 没有 问题
原文地址：https://www.cnblogs.com/caijiaming/p/10387166.html
