题意:给定n个组,每个组包括a,b两个值,问除去k个不选,能得到的
100∗∑ai∑bi
的最大值是多少。
思路:二分这个东西
∑ai∑bi
,然后判断即可
二分的单调性:这个很明显,如果能得到结果满足
>=x
,那么对于所有的
xi<x
一定有值满足大于等于
xi
(废话)
二分的check(mid):要满足
∑ai∑bi>=mid
则
∑(ai−mid∗bi)>=0
,因为有k个不用,对这个值排个序,得到的结果从大到小选,与0比较即可。
#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<cstdlib>
#include<list>
#include<stack>
#include<cmath>
#include<iomanip>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
void debug() {cout << "ok running!" << endl;}
int n, k;
struct node
{
int v, w, id;
double x;
bool operator < (node& b)
{
return x > b.x;
}
} a[100005];
bool check(double mid)
{
for(int i = 0; i < n; ++i)
a[i].x = a[i].v - mid*a[i].w;
sort(a, a+n);
double sum = 0;
for(int i = 0; i < n-k; ++i)
sum += a[i].x;
if(sum >= 0)
return 1;
else return 0;
}
int main()
{
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
#endif // ONLINE_JUDGE
while(cin >> n >> k)
{
if(n == 0 && k == 0)
break;
for(int i = 0; i < n; ++i)
{
a[i].id = i+1;
cin >> a[i].v;
}
for(int i = 0; i < n; ++i)
cin >> a[i].w;
double l = 0.0, r = 999999999.0;
while(r - l > 1e-4)
{
double mid = (l+r) / 2;
//cout << mid << endl;
if(check(mid))
l = mid;
else r = mid;
}
cout << (int)((l+r)/2*100+0.5) << endl;
}
return 0;
}