1.题目描述：

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

2.题意概述：

0-1分数规划+二分

``````#include<bits/stdc++.h>
#define INF 0x7fffffff
#define esp 1e-15
using namespace std;
const int maxn=1e9+10;
long long n,k;
double a[maxn],b[maxn],y[maxn];
int judge(double x){
for(int i=0;i<n;i++)
y[i]=a[i]-x*b[i];
sort(y,y+n,greater<double>());
double sum=0;
for(int i=0;i<n-k;i++)
sum+=y[i];
return sum>0;
}
int main(){
while(~scanf("%lld%lld",&n,&k),n|k){
for(int i=0;i<n;i++)
scanf("%lf",&a[i]);
for(int i=0;i<n;i++)
scanf("lf",&b[i]);
double l=0,r=INF;
double mid;
while(r-l>esp){
mid=l+(r-l)/2;
if(judge(mid))
l=mid;
else
r=mid;
}
printf("%.0lf\n", mid * 100);
}
return 0;
}``````
``````#include<bits/stdc++.h>
#define INF 0x7fffffff
#define esp 1e-15
using namespace std;
const int maxn=1e9+10;
long long n,k;
double a[maxn],b[maxn];
double Dinkelbach(double L){
double y[maxn], suma,sumb , minu;
bool vis[maxn] ;
int i , j , index;
memset(vis,false,sizeof(vis));
suma = sumb =0;
for(i = 1; i <=n ;i++)y[i] = a[i] - L*b[i];
for(i = 1;i <=n-k ; i++){
minu = -INF;
for(j = 1;j <=n ;j++){
if(!vis[j]&&y[j] >minu){
index = j;
minu = y[j];
}
}
vis[index] = true;
suma+=a[index];sumb+=b[index];
}
return suma/sumb;
}
int main(){
while(~scanf("%lld%lld",&n,&k),n|k){
for(int i=0;i<n;i++)
scanf("%lf",&a[i]);
for(int i=0;i<n;i++)
scanf("lf",&b[i]);
double L1,L2;
L1=0;
while(1){
L2=Dinkelbach(L1);
if(L1==L2)break;
L1=L2;
}
printf("%.0lf\n",L1*100);
}
return 0;
}
``````