The World Programming Olympics Medal is a metal disk, consisting of two parts: the first part is a ring with outer radius of r1 cm, inner radius of r2 cm, (0?<?r2?<?r1) made of metal with density p1 g/cm3. The second part is an inner disk with radius r2 cm, it is made of metal with density p2 g/cm3. The disk is nested inside the ring.

The Olympic jury decided that r1 will take one of possible values of x1,?x2,?…,?xn. It is up to jury to decide which particular value r1 will take. Similarly, the Olympic jury decided that p1 will take one of possible value of y1,?y2,?…,?ym, and p2 will take a value from list z1,?z2,?…,?zk.

According to most ancient traditions the ratio between the outer ring mass mout and the inner disk mass min must equal , where A,?B are constants taken from ancient books. Now, to start making medals, the jury needs to take values for r1, p1, p2 and calculate the suitable value of r2.

The jury wants to choose the value that would maximize radius r2. Help the jury find the sought value of r2. Value r2 doesn’t have to be an integer.

Medal has a uniform thickness throughout the area, the thickness of the inner disk is the same as the thickness of the outer ring.

Input
The first input line contains an integer n and a sequence of integers x1,?x2,?…,?xn. The second input line contains an integer m and a sequence of integers y1,?y2,?…,?ym. The third input line contains an integer k and a sequence of integers z1,?z2,?…,?zk. The last line contains two integers A and B.

All numbers given in the input are positive and do not exceed 5000. Each of the three sequences contains distinct numbers. The numbers in the lines are separated by spaces.

Output
Print a single real number — the sought value r2 with absolute or relative error of at most 10?-?6. It is guaranteed that the solution that meets the problem requirements exists.

Examples
Input
3 1 2 3
1 2
3 3 2 1
1 2
Output
2.683281573000
Input
4 2 3 6 4
2 1 2
3 10 6 8
2 1
Output
2.267786838055
Note
In the first sample the jury should choose the following values: r1?=?3, p1?=?2, p2?=?1.
說了一大堆,就是一個大盤子中間挖去了一塊,然后呢,這兩個的密度還不同.已經(jīng)給你了r1,p1,p2,(r1就是盤子的直徑),讓你去求r2(內(nèi)圓的直徑).要求是內(nèi)外的比值一定要等于A/B.就是推公式…會得到r2^2=(Bp1r1r1)/(Ap2+Bp1).開始我看到這里的時候,我就想,暴力一發(fā),然后就t了,這么做復(fù)雜度是O(mnk),肯定會爆的.然后繼續(xù)看式子.上下同時除以p1不就好了嗎,這么做復(fù)雜度是O(nm).就可以啦.
代碼如下:

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std;const int maxx=5e3+10; double r[maxx]; double p1[maxx]; double p2[maxx]; double a,b; int n,m,k;int main() {scanf("%d",&n);for(int i=0;i<n;i++) scanf("%lf",&r[i]);scanf("%d",&m);for(int i=0;i<m;i++) scanf("%lf",&p1[i]);scanf("%d",&k);for(int i=0;i<k;i++) scanf("%lf",&p2[i]);scanf("%lf%lf",&a,&b);double maxn=99999999;for(int i=0;i<k;i++){for(int j=0;j<m;j++){maxn=min(maxn,p2[i]/p1[j]);}}double m=-1;for(int i=0;i<n;i++){double x=sqrt(b*r[i]*r[i]/(a*maxn+b));if(x>m){m=x;}}printf("%.12lf\n",m); }

努力加油a啊,(^o)/~

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