POJ_P2976 Dropping tests(01分数规划)

POJ传送门
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8729 Accepted: 3041
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.

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
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

随便搞搞了解一下，了解算法
一个二分一个Dinkelbach
速度比较(第一个是二分，第二个是Dinkelbach)

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1005
#define eps 1e-6
inline int in(int x=0,int v=1,char ch=getchar()){
    while(ch!='-'&&(ch>'9'||ch<'0')) ch=getchar();if(ch=='-') v=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x*v;
}
struct str{double d;int num;}d[N];
int a[N],b[N];int n,m,k;
inline int cmp(str a,str b){return a.d>b.d;}
void work1(){//二分 
    m=n-k;double l,r,mid,tmp;mid=0;
    for(int i=1;i<=n;i++) if(1.0*a[i]/b[i]>mid) mid=1.0*a[i]/b[i];
    l=0,r=mid;
    while(fabs(r-l)>eps){
        mid=(l+r)/2.0;
        for(int i=1;i<=n;i++) d[i].d=a[i]-mid*b[i],d[i].num=i;
        sort(d+1,d+n+1,cmp);tmp=0;
        for(int i=1;i<=m;i++) tmp+=d[i].d;
        if(tmp>0) l=mid;else r=mid;
    }
    printf("%.0f\n",l*100);
}
void work2(){//Dinkelbach
    m=n-k;long long p,q;double ans=0,L=1;
    while(fabs(ans-L)>eps){
        ans=L;
        for(int i=1;i<=n;i++) d[i].d=a[i]-L*b[i],d[i].num=i;
        sort(d+1,d+n+1,cmp);p=q=0;
        for(int i=1;i<=m;i++) p+=a[d[i].num],q+=b[d[i].num];
        L=p*1.0/q;
    }
    printf("%.0f\n",L*100);
}
int main(){
    while(scanf("%d%d",&n,&k)!=EOF&&(n||k)){
        for(int i=1;i<=n;i++) a[i]=in();
        for(int i=1;i<=n;i++) b[i]=in();
        work1();
        work2();
    }
    return 0;
}