POJ 2976:Dropping tests 01 分数规划

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12291 Accepted: 4293

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).

POJ竟然用不了读入优化。。。努WA好多发

01分数规划裸题

#include<cmath>#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<complex>#include<iostream>#include<algorithm>#include<iomanip>#include<vector>#include<string>#include<queue>#include<set>#include<map>using namespace std;typedef double db;inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}const db eps=1e-6;const int N=1010;int n,m;db d[N],a[N],b[N];db work(db x){for(int i=1;i<=n;i++)d[i]=a[i]-x*b[i];sort(d+1,d+n+1);db sum=0.0;for(int i=m+1;i<=n;i++)sum+=d[i];return sum;} int main(){while(scanf("%d%d",&n,&m),n||m){for(int i=1;i<=n;i++)scanf("%lf",&a[i]);for(int i=1;i<=n;i++)scanf("%lf",&b[i]);db l=0.0,r=1.0;while(fabs(r-l)>eps){db mid=(l+r)/2;if(work(mid)>0)l=mid;else r=mid;}printf("%.0lf\n",l*100);}return 0;}/*3 15 0 25 1 64 21 2 7 95 6 7 90 083100*/