POJ 2976

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5459		Accepted: 1889

Description

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 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).

Source

Stanford Local 2005

 

二分平均值。

 

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 
 6 using namespace std;
 7 
 8 #define maxn 1005
 9 #define eps 1e-6
10 
11 int n,k;
12 int a[maxn],b[maxn];
13 
14 bool judge(double x) {
15         double sum = 0;
16         double f[maxn];
17         for(int i = 1; i <= n; ++i) {
18                 f[i] = a[i] - x * b[i];
19         }
20 
21         sort(f + 1,f + n + 1);
22 
23         for(int i = n; i >= n - (n - k) + 1; --i) {
24                 sum += f[i];
25         }
26 
27         return sum >= 0;
28 
29 }
30 
31 void solve() {
32         double l = 0,r = 0;
33 
34         for(int i = 1; i <= n; ++i) {
35                 r += a[i];
36         }
37 
38         while(r - l >= eps) {
39                 double mid = (l + r) / 2;
40                 if(judge(mid)) l = mid;
41                 else r = mid;
42 
43         }
44 
45        // printf(" l = %f\n",l);
46 
47         printf("%d\n",(int)(100 * r + 0.5));
48 }
49 
50 int main() {
51        、、 freopen("sw.in","r",stdin);
52 
53         while(~scanf("%d%d",&n,&k)) {
54         if(!n && !k) break;
55         for(int i = 1; i <= n; ++i) {
56                 scanf("%d",&a[i]);
57         }
58 
59         for(int i = 1; i <= n; ++i) {
60                 scanf("%d",&b[i]);
61         }
62 
63         solve();
64 
65         }
66 
67         return 0;
68 }

View Code

 

转载于:https://www.cnblogs.com/hyxsolitude/p/3611868.html