∑ s [ i ] ? c [ i ] ∑ s [ i ] > = t a r ∑ s [ i ] ? c [ i ] > = t a r ? ∑ s [ i ] ; 然 后 把 每 一 项 变 成 相 应 的 形 式 就 行 b [ i ] = s [ i ] ? c [ i ] ? s [ i ] ? t a r 然 后 贪 心 的 去 除 至 多 k 项 的 负 数 \frac{\sum{s[i]*c[i]}}{\sum{s[i]}} >= tar\\ \sum{s[i]*c[i]} >= tar * \sum{s[i]};\\ 然后把每一项变成相应的形式就行\\ b[i] = s[i] * c[i] - s[i] * tar\\ 然后贪心的去除至多k项的负数 ∑s[i]∑s[i]?c[i]?>=tar∑s[i]?c[i]>=tar?∑s[i];然后把每一项变成相应的形式就行b[i]=s[i]?c[i]?s[i]?tar然后贪心的去除至多k项的负数
// Problem: gpa
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/22353/T
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-04-30 11:32:20
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> Vll;
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
ll ans = 0;
while (y) {
if (y & 1)
ans = (ans + x) %mod;
y >>= 1;
x = (x + x) %mod;
}
return ans;
}
inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
// inline ll inv (ll a) {
// return qmi(a, mod - 2);
// }
const int N = 1e5 +10;
int n, k;
double c[N], s[N];
double b[N];
bool check(double tar) {
for (int i = 1; i <= n;i ++)
b[i] = s[i] * c[i] - tar * s[i];
sort(b + 1, b + 1 + n);
double ans = 0;
int id = k + 1;
for (int i = 1; i <= k; i ++) {
if (b[i] >= 0) {
id = i;
break;
}
}
for (int i = id; i <= n;i ++)
ans += b[i];
if (ans >= 0) return 1;
return 0;
}
void solve() {
cin >> n >> k;
for (int i = 1; i <= n; i ++)
cin >> s[i];
for (int i = 1; i <= n; i ++) {
cin >> c[i];
}
double l = 0, r = 1e10;
for (int i = 0; i <= 100; i ++) {
double mid = (l + r) /2;
if (check(mid)) l = mid;
else r = mid;
}
printf("%.12lf\n", l);
}
int main () {
// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}
