Acm:how many integers can you find-the topic of number theory-the simple application of the repulsion principle +GCD

How many integers can you findTime limit:5000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U DescriptionNow you get a number N, and a m-integers set, you should find out how many integers which is small than N, that they can Divided exactly by any integers in the set. For example, n=12, and M-integer set are {2,3}, so there are another set {2,3,4,6,8,9,10}, all the integers of the set can B e divided exactly by 2 or 3. As a result, you just output the number 7.

Input

There is a lot of cases. For each case, the first line contains integers N and M. The follow line contains the M integers, and all of the them is different from each other. 0<n<2^31,0<m<=10, and the M integer is non-negative and won ' t exceed 20.

Output

For each case, output the number.

Sample Input

12 22 3

Sample Output

7

/* /test Instructions: gives the number of N and M input m, finds a multiple of all m numbers and, in multiples of MI, is less than n, and outputs the total number of all numbers. If a number is at the same time a multiple of three numbers, a number of times the number of multiples of C (3,1) =3 two times the number of times C (3,2) =3 three number of times the number is C (3,3) =13-3+1=1, only one time, and so on one count is 5 number of multiples C (5,1) =5c (5,2) =10C (5,3) =10c (5,4) =5c (5,5) =15-10+10-5+1=1 six numbers C (6,1) =6c (6,2) =15c (6,3) =20c (6,4) =15c (6,5) =6c (6,6) =16-15+20-15+6-1= 1 then because the number is not more than 10, you can use the idea of enumerating subsets to do this topic. So use DFS. Finally, there is a place to pay attention is in the DFS inside to determine the product here, to use GCD, at first did not think of a sample. AC Code:/*

#include "map" #include "Cmath" #include "string" #include "Cstdio" #include "vector" #include "CString" #include "iostream "#include" algorithm "using namespace std;typedef long LL; LL A[15];int n,m,cnt; LL ans,x; ll GCD (ll A,ll b) {return B?GCD (b,a%b): A;} void DFS (int x,ll axb,int num) {AXB=A[X]/GCD (A[X],AXB) *axb;if (num&1) ans+= (n-1)/axb;else ans-= (n-1)/axb;//cout << "Now ans is:" <<ans<<endl;  Check for (int i=x+1; i<cnt; i++) DFS (i,axb,num+1);} int main () {while (~scanf ("%d%d", &n,&m)) {ans=0;cnt=0;for (int i=0; i<m; i++) {scanf ("%i64d", &x); if (x!=0 ) A[cnt++]=x;} for (int i=0; i<cnt; i++) {DFS (i,a[i],1);       Use DFS to enumerate the circumstances of each selection. }printf ("%d\n", ans); return 0;}

　　

Acm:how many integers can you find-the topic of number theory-the simple application of the repulsion principle +GCD