UVA 10050 Hartals (罷工指數，暴力枚舉。)

 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).

	1	2	3	4	5	6	7	8	9	10	11	12	13	14
Days
	Su	Mo	Tu	We	Th	Fr	Sa	Su	Mo	Tu	We	Th	Fr	Sa
Party 1			x			x			x			x
Party 2				x				x				x
Party 3								x
Hartals			1	2				3	4			5

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

題目大意：

給你一個連續的天數，在星期六和星期五是休息，罷工是沒用的。再給你P個組織，每個組織給你一個數字，表示在這個數字的倍數上此政黨會罷工。問你這些天內罷工的總天數。

解題思路：

暴力枚舉，注意，可能多個組織在同一天罷工，這隻算一天。

代碼：

#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;
}