poj 2976(01分数规划最基础)

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6151 Accepted: 2131

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

题目的意思是，给你n对数(a,b)求sigma(a[i]) / sigma(b[i]) 的最大值，其中需要去掉K个数，也就是说求其中N-K个数的最大值。

这个题是01分数规划的超级裸题，提供两种方法，二分和一个叫不上名字的。后者要快，但是不精确，具体用吧。详见代码。

二分：

#include <iostream>#include <cstdio>#include <algorithm>#define MAXN 1111#define eps 1e-6int a[MAXN], b[MAXN];double score[MAXN];using namespace std;int main(){    int n, k;    while (~scanf("%d%d", &n, &k) && (n || k)) {        for (int i = 0; i < n; i++)            scanf("%d", &a[i]);        for (int i = 0; i < n; i++)            scanf("%d", &b[i]);        double l = 0.0, r = 1.0;        while (l < r - eps) {            double ans = 0;            double mid = (l + r) / 2.0;            for(int i = 0; i < n; i++)                score[i] = a[i] - mid * b[i];            sort(score, score + n);            for (int i = n - 1; i >= k; i--)                ans += score[i];            if(ans >= 0.0) l = mid;            else r = mid;        }        printf("%d\n",(int)(l * 100 + 0.5));    }    return 0;}

方法二：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;struct N{    int a, b;    double c;}node[1111];bool cmp(N a, N b) {    return a.c > b.c;}int main(){    int n, k;    while (~scanf("%d%d", &n, &k) && (n + k) != 0) {        for (int i = 0; i < n; i++) {            scanf("%d", &node[i].a);        }        for (int i = 0; i < n; i++) {            scanf("%d", &node[i].b);        }        double l = 0.5, ans = 0;        while (fabs(l - ans) >= 1e-6) {            ans = l;            for (int i = 0; i < n; i++)                node[i].c = node[i].a - l * node[i].b;            sort(node, node + n, cmp);            double suma = 0, sumb = 0;            for (int i = 0; i < n - k; i++) {                suma += node[i].a;                sumb += node[i].b;            }            l = suma / sumb;        }        printf("%d\n", (int)(l * 100 + 0.5));    }    return 0;}