POJ - 2976 Dropping tests-白红宇

POJ - 2976 Dropping tests

阅读量：5268 次

发布时间：2019-06-14

本文共 2264 字，大约阅读时间需要 7 分钟。

In a certain course, you take n tests. If you get a_i out of b_i 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 a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

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).

二分基础题

1 #include 
       
         2 using namespace std; 3 #include
        
          4 #include
         
           5 #include
          
            6 #include
           
             7 #include
            
              8 #include
              9 #include
              
               10 #include
               
                11 #include
                
                 12 #include
                 
                  13 #include 
                  
                   14 #include 
                   
                    15 #include
                    
                     16 #include
                     
                      17 double a[1100];18 double b[1100];19 double c[1100];20 const double MIN=1e-7;21 int n,m;22 int main()23 {24 while(cin>>n>>m)25 {26 if(n==0&&m==0)27 break;28 for(int i=1;i<=n;i++)29 cin>>a[i];30 for(int i=1;i<=n;i++)31 cin>>b[i];32 double kaishi=0.0,jieshu=1.0;33 double mid;34 while(jieshu-kaishi>MIN)35 {36 mid=(kaishi+jieshu)/2.0;37 //cout<
                      
                       <<"_"<
                       
                        =0)45 kaishi=mid;46 else47 jieshu=mid;48 }49 int qqq=mid*1000;50 if(qqq%10>=5)51 qqq+=10;52 qqq/=10;53 cout<
                        
                         <

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转载于:https://www.cnblogs.com/dulute/p/7966690.html

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