POJ2976-Dropping tests

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12192 Accepted: 4260

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 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 decision boundary (i.e., you can assume that the average is never 83.4997).

Source

Stanford Local 2005

题意：n场考试中分别答对a_i题，总题数分别为b_i，允许去掉k场考试，求能达到的最高准确率
解题思路：这题不能用单位价值的排序来求答案，而是需要二分准确率在当前准确率为x的情况下，这场考试“额外”对的题目数量是a_i – x * b_i，当然这个值有正有负，恰好可以作为“贡献度”的测量。于是利用这个给考试排个降序，刷掉后k个

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n,k;struct node{int a, b;double p;}x[1009];bool cmp(node a, node b){return a.p > b.p;}bool check(double xx){for (int i = 1; i <= n; i++) x[i].p = 1.0*x[i].a -1.0* x[i].b*xx;sort(x + 1, x + 1 + n, cmp);double sum1 = 0, sum2 = 0;for (int i = 1; i <= n - k; i++) sum1 += x[i].a, sum2 += x[i].b;return sum1 / sum2 > xx;}int main(){while (~scanf("%d%d", &n, &k)){if (!n && !k) break;for (int i = 1; i <= n; i++) scanf("%d", &x[i].a);for (int i = 1; i <= n; i++) scanf("%d", &x[i].b);double l = 0, r = 1;while (fabs(r - l) > 1e-8){double mid = (l + r) / 2;if (check(mid)) l = mid;else r = mid;}printf("%.0f\n",l*100);}return 0;}