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 taht D 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. All M[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 exceed 10^- 4.

Sample Input

Input

1
2
0

Output

0.500000

Input

2
2 3
1 0

Output

0.666667

題意：就是給你兩個數組，每個數組包含N個數，讓你來求x%a[i] == b[i]的概率事多大，

思路；直接暴力解決;枚舉x來求概率，把概率放大來求保留小數點後六位，那麼放大後就是在10^7方來找x

AC代碼：

#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;
}