codeforces 656b Scrambled



Btoh yuo adn yuor roomatme lhoate wianshg disehs, btu stlil sdmoeboy msut peorrfm tihs cohre dialy. Oen dya yuo decdie to idourtcne smoe syestm. Yuor rmmotaoe sstgegus teh fooniwllg dael. Yuo argee on tow arayrs of ientgres M adn R, nmebur upmicnog dyas (induiclng teh cunrret oen) wtih sicsescuve irnegets (teh ceurrnt dya is zreo), adn yuo wsah teh diehss on dya D if adn olny if terhe etsixs an iednx i scuh tahtD mod M[i] = R[i], otwsehrie yuor rmootmae deos it. Yuo lkie teh cncepot, btu yuor rmotaome's cuinnng simle meaks yuo ssecupt sthnoemig, so yuo itennd to vefriy teh fnerisas of teh aemnrgeet.

Yuo aer geivn ayarrs M adn R. Cuaclatle teh pceanregte of dyas on wchih yuo edn up dnoig teh wisahng. Amsuse taht yuo hvae iiiftlneny mnay dyas aehad of yuo.

Input

The first line of input contains a single integer N (1 ≤ N ≤ 16).

The second and third lines of input contain N integers each, all between 0 and 16, inclusive, and represent arrays M and R, respectively. AllM[i] are positive, for each iR[i] < M[i].

Output

Output a single real number. The answer is considered to be correct if its absolute or relative error does not exceed10^- 4.

Sample Input

Input

120

Output

0.500000

Input

22 31 0

Output

0.666667

题意：就是给你两个数组，每个数组包含N个数，让你来求x%a[i] == b[i]的概率事多大，

思路；直接暴力解决;枚举x来求概率，把概率放大来求保留小数点后六位，那么放大后就是在10^7方来找x

AC代码：

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <set>#include <map>using namespace std;int n;int a[50];int b[50];int main() {    cin >> n;    for(int i = 0;i < n;i++){        cin >> a[i];    }    for(int i = 0;i < n;i++){        cin >> b[i];    }    double ans = 0;    for(int i = 1;i < 10000000;i++){        for(int j = 0;j < n;j++){            if(i%a[j] == b[j]){                ans += 1;                break;            }        }    }    printf("%.6f\n",ans/10000000);    return 0;}