POJ 2976/ZOJ 3068 Dropping tests 01 score planning, pojzoj

POJ 2976/ZOJ 3068 Dropping tests 01 score planning, pojzoj

Card accuracy 01 score planning...

Dropping tests

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:6464		Accepted:2230

Description

In a certain course, you takeNTests. If you getAiOutBiQuestions correct on testI, Your cumulative average is defined to be

.

Given your test scores and a positive integerK, Determine how high you can make your cumulative average if you are allowed to drop anyKOf 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 containsNIntegers indicatingAiFor allI. The third line containsNPositive integers indicatingBiFor allI. It is guaranteed that 0 ≤Ai≤Bi≤ 1,000,000,000. The end-of-file is marked by a test caseN=K= 0 and shoshould not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after droppingKOf the given test scores. The average shocould be rounded to the nearest integer.

Sample Input

3 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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 demo-boundary (I. e ., you can assume that the average is never 83.4997 ).

Source

Stanford Local 2005

# Include <iostream> # include <cstring> # include <cstdio> # include <algorithm> using namespace std; const int maxn = 1111; const double eps = 1e-8; int n, k; double a [maxn], B [maxn]; double d [maxn]; bool check (double L) {for (int I = 0; I <n; I ++) {d [I] = a [I]-B [I] * L;} sort (d, d + n); double sum = 0 .; for (int I = k; I <n; I ++) {sum + = d [I];} if (sum> eps) return true; return false ;} int main () {while (scanf ("% d", & n, & k )! = EOF) {if (n = 0 & k = 0) break; for (int I = 0; I <n; I ++) scanf ("% lf", a + I); for (int I = 0; I <n; I ++) scanf ("% lf", B + I ); double low = 0, high = 1, mid, ans; while (low + eps