POJ 2976 Dropping tests 0/1分数规划问题 最大化平均值 贪心+二分

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 n positive 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).

方法：可行函数单调递减，想到用二分快速搜索，可行函数是线性的，满足贪心原理

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;const double eps = 1e-4;int a[1005], b[1005];double f[1005];int n, k;bool cmp(double i, double j){    return i > j;}bool C(double v){    for(int i=0; i<n; i++)        f[i] = a[i] - v*b[i];    sort(f, f+n, cmp);    double sum = 0;    for(int i=0; i<n-k; i++)        sum += f[i];    if(sum >= 0) return 1;    else return 0;}int main(){    while(scanf("%d %d", &n, &k) == 2){        if(n==0 && k==0) break;        for(int i=0; i<n; i++)            scanf("%d", &a[i]);        for(int i=0; i<n; i++)            scanf("%d", &b[i]);        double lb = 0, ub = 1.0;        while(ub - lb > eps){            double mid = (lb+ub)/2;            if(C(mid)) lb = mid;            else ub = mid;        }        printf("%.0lf\n", lb*100);    }    return 0;}