Hartals

A social research organization has determined a simple set of parameters to simulate the behavior
of the political parties of our country. One of the parameters is a positive integerh (called
the hartal parameter)
that denotes the average number of days between two successive hartals (strikes)
called by the corresponding party. Though the parameter is far too simple to be flawless, it can still be used to forecast the damages caused by hartals.
The following example will give you a clear idea:

Consider three political parties. Assume h1 = 3, h2 =
4 and h3 = 8 where hi is
the hartal parameter for party i ( i = 1, 2, 3). Now, we will simulate the behavior of these three parties for N = 14 days. One must always start the simulation on a Sunday
and assume that there will be no hartals on weekly holidays (on Fridays and Saturdays).

The simulation above shows that there will be exactly 5 hartals (on days 3, 4, 8, 9 and 12) in 14 days. There will be no hartal on day 6 since it is a Friday. Hence we lose 5 working days in 2 weeks.

In this problem, given the hartal parameters for several political parties and the value of N, your job is to determine the number of working days we lose in those N days.

Input

The first line of the input consists of a single integer T giving the number of test cases to follow.

The first line of each test case contains an integer N ( )
giving the number of days over which the simulation must be run. The next line contains another integer P( )
representing the number of political parties in this case. The ith of the next P lines contains a positive integer hi (which
will never be a multiple of 7) giving thehartal parameter for party i ( ).

Output

For each test case in the input output the number of working days we lose. Each output must be on a separate line.

Sample Input

2
14
3
3
4
8
100
4
12
15
25
40

Sample Output

5
15

题目大意：

给你一个连续的天数，在星期六和星期五是休息，罢工是没用的。再给你P个组织，每个组织给你一个数字，表示在这个数字的倍数上此政党会罢工。问你这些天内罢工的总天数。

解题思路：

暴力枚举，注意，可能多个组织在同一天罢工，这只算一天。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int main(){
    int t,num,p[110];
    bool visited[5000];
    cin>>t;
    while(t--){
        int cnt=0,nump;
        cin>>num>>nump;
        memset(visited,false,num+1);
        for(int i=0;i<nump;i++) cin>>p[i];
        for(int i=0;i<nump;i++){
            for(int j=1;j<=num;j++){
                if(j%p[i]==0&&j%7!=6&&j%7!=0&&!visited[j]){
                    cnt++;
                    visited[j]=true;
                }
            }
        }
       cout<<cnt<<endl;
    }
    return 0;
}