POJ 2976 Dropping tests 01分数规划Dinkelbach算法

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13732 Accepted: 4817

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 npositive 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

01分数规划入门，使用Dinkelbach算法，在现有的可行解下，不断迭代得到最优解。

#include <cstdio>#include <iostream>#include <algorithm>#include <math.h>#include <cmath>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=10005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const db eps=1e-6;struct Pack{db a,b,d;};Pack p[maxn];bool cmp(Pack x,Pack y) {return x.d<y.d;}db Dinkelbach(int n,int k) {db ans=0,l;while (true) {l=ans;int i; db s0,s1;for (i=1;i<=n;i++) {p[i].d=p[i].a-l*p[i].b;}sort(p+1,p+n+1,cmp);s0=s1=0;for (i=k+1;i<=n;i++) {s0+=p[i].a;s1+=p[i].b;}ans=s0/s1;if (fabs(l-ans)<eps) return ans;}}int main() {int n,k;scanf("%d%d",&n,&k);while (n||k) {int i;ll s1,s0;s1=s0=0;for (i=1;i<=n;i++)scanf("%lf",&p[i].a);for (i=1;i<=n;i++) scanf("%lf",&p[i].b);db ans=Dinkelbach(n,k);printf("%.0lf\n",ans*100.0);scanf("%d%d",&n,&k);}return 0;}