The strength value of the small p

Time limit for strength of small p: 10000ms single point time: 1000ms memory limit: 256MB description

Little hi in the virtual world has a small pet small p. The small P has K attributes, and the initial value of each property is AI. Little ho gave small hi a few pills, each pill can raise the small p specify the attribute 1 points. With the property value, we can calculate the strength value of the small P = (C1 (1/B1)) * (C2 (1/B2)) *...* (CK (1/BK)), where CI is the final value of the small P-I property (ai+ pill added attribute). Small Ho is known to give small hi pills a total of n, ask the strength of the small p max can reach?

Input

The first line contains two integer n,k, each representing the number of pills and the number of attributes.

The second behavior is k integer a1-ak, meaning as described earlier.

The third act K integer b1-bk, meaning as described above.

For 30% of data, meet 1<=n<=10, 1<=k<=3

For 100% of data, meet 1<=n<=100000, 1<=k<=10

For 100% of data, meet 1<=ai<=100, 1<=bi<=10

Output

The highest strength value the output small p can achieve.

As long as the relative or absolute error between your results and the correct answer does not exceed 1 per thousand, it is considered the correct output.

Sample input

5 21 13 2

Sample output

2.88
Analysis: After the answer to log found that in fact multiplication into an addition, and then the priority queue simulation can be;
Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<climits>#include<cstring>#include<string>#include<Set>#include<map>#include<queue>#include<stack>#include<vector>#include<list>#defineRep (I,m,n) for (i=m;i<=n;i++)#defineRSP (It,s) for (Set<int>::iterator It=s.begin (); It!=s.end (); it++)#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineVI vector<int>#definePB Push_back#defineMP Make_pair#defineFi first#defineSe Second#definell Long Long#definePi ACOs (-1.0)#definePII pair<int,int>Const intmaxn=1e5+Ten;Const intdis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};using namespacestd;ll gcd (ll p,ll q) {returnq==0? P:GCD (q,p%q);} ll Qpow (ll p,ll q) {ll F=1; while(q) {if(q&1) f=f*p%mod;p=p*p%mod;q>>=1;}returnF;}intn,m,k,t;Doubleans;structnode{intX,y,id; BOOL operator< (ConstNODE&AMP;P)Const    {        return 1.0/y* (log (x+1)-log (x)) <1.0/p.y* (log (p.x+1)-log (p.x)); }}A[MAXN];p riority_queue<node>p;intMain () {inti,j; scanf ("%d%d",&n,&k); Rep (I,0, K-1) scanf ("%d", &a[i].x), a[i].id=i; Rep (I,0, K-1) scanf ("%d",&a[i].y); Rep (I,0, K-1) P.push (A[i]); Rep (I,0, N-1) {node Q=P.top ();        P.pop (); q.x++,a[q.id].x++;    P.push (q); } ans=1.0; Rep (I,0, K-1) Ans=ans*pow (a[i].x,1.0/a[i].y); printf ("%.7f\n", ans); //System ("pause");    return 0;}

The strength value of the small p