poj 2976 Dropping tests
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4559 | Accepted: 1542 |
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.
//0-1 分数规划
//详见: http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html
#include <iostream> #include <math.h> #include <vector> #include <queue> #include <stack> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int N=1010; const double eps=1e-6; double a[N],b[N]; struct node { double z; int i; bool operator<(const node &b) const { return z<b.z; } }oz[N]; int main() { int n,k; int i; while(scanf("%d %d",&n,&k)==2&&n|k) { for(i=0;i<n;i++) scanf("%lf",&a[i]); for(i=0;i<n;i++) scanf("%lf",&b[i]); double ans,temp=0; double p,q; while(1) { ans=temp; for(i=0;i<n;i++) { oz[i].i=i; oz[i].z=a[i]-temp*b[i]; } sort(oz,oz+n); p=q=0; for(i=k;i<n;i++) { p+=a[oz[i].i]; q+=b[oz[i].i]; } temp=p/q; if(fabs(temp-ans)<eps) break; } printf("%.0lf\n",ans*100); } return 0; }