How many integers can you find
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.
Input
There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.
Output
For each case, output the number.
Sample Input
12 2 2 3
Sample Output
7
/*/ 题意： 给出N和M 输入M个数，找出所有M个数的倍数并且，Mi的倍数小于N，输出所有数的总个数。 如果一个数同时是三个数的倍数 单独记一个数的倍数次数为C（3，1） =3 记两个数的倍数次数为 C（3，2）=3 记三个数的倍数次数为 C（3，3）=1 3-3+1=1，只记一次依次类推 一个数为5个数的倍数 C（5，1）=5 C（5，2）=10 C（5，3）=10 C（5，4）=5 C（5，5）=1 5-10+10-5+1=1 六个数 C（6，1）=6 C（6，2）=15 C（6，3）=20 C（6，4）=15 C（6，5）=6 C（6，6）=1 6-15+20-15+6-1=1 然后因为数字不超过10个，可以运用枚举子集的思想去做这个题目。 所以用到DFS。 最后有一个地方要注意就是在DFS里面判断积这里，要用GCD，一开始没想到过不了样例。 AC代码： /*/
#include"map"
#include"cmath"
#include"string"
#include"cstdio"
#include"vector"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long 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; //检查
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); //用DFS去枚举每种选择的情况。
}
printf("%d\n",ans);
}
return 0;
}