学步园

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 integer
h (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 h_i 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
h_i (which will never be a multiple of 7) giving the
hartal 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

#include <stdio.h>
void main()
{int t,i,j,n,m,h,sum,a[4000];
 scanf("%d",&t);
 while (t--)
 {
 scanf("%d",&n);
 for (i=1;i<=n;i++)
 a[i]=0; sum=0;
 scanf("%d",&m);
 for (i=1;i<=m;i++)
 {scanf("%d",&h);
  for (j=1;j*h<=n;j++)
  a[j*h]=1;
 }
 for (i=0;7*i<=n;i++)
 {a[7*i]=0;a[7*i+6]=0;}
 for (i=1;i<=n;i++)
 sum+=a[i];
 printf("%d\n",sum);
 }
}