POJ 2976 Dropping tests

Dropping tests

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2976
64-bit integer IO format: %lld      Java class name: Main

 

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 nintegers 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

Source

Stanford Local 2005

 

解题：01分数规划。Dinkelbach

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const double exps = 1e-6;
struct node{
    double c,a,b;
};
node p[1010];
int n,k;
bool cmp(const node &x,const node &y){
    return x.c < y.c;
}
double test(){
    double ans,x,y,tmp = 0;
    while(true){
        ans = tmp;
        for(int i = 0; i < n; i++)
            p[i].c = p[i].a - ans*p[i].b;
        sort(p,p+n,cmp);
        x = y = 0;
        for(int i = k; i < n; i++){
            x += p[i].a;
            y += p[i].b;
        }
        tmp = x*1.0/y;
        if(fabs(tmp - ans) < exps) return ans;
    }
}
int main() {
    while(scanf("%d %d",&n,&k),n||k){
        for(int i = 0; i < n; i++)
            scanf("%lf",&p[i].a);
        for(int i = 0; i < n; i++)
            scanf("%lf",&p[i].b);
        printf("%.0f
",test()*100);
    }
    return 0;
}