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
能被给出的第二行的数整除的数的个数；
sum=被一个整除-被两个整除+被三个整除-。。。。。。
注意一个他自己本身不算
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#pragma comment(linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define EPS 1e-6
#define INF (1<<24)
using namespace std;
int m,n,cnt;
int num[20];
int visi[20];
long long int sum;
long long int gcd(long long int a,long long int b) //最大公约数
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
long long int lcm(long long int a,long long int b) //最小公倍数
{
return a/gcd(a,b)*b;
}
void solve()
{
int i,flag=0;//记录有多少个数
long long int t=1,ans;
for(i=0;i<cnt;i++)
{
if(visi[i])
{
flag++;
t=lcm(t,num[i]);
}
}
ans=n/t;
if(n%t==0) ans--;
if(flag%2==1) sum=sum+ans; //奇数加，偶数减
else sum=sum-ans;
}
int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
int i,j;
int a;
cnt=0;
sum=0;
for(i=0;i<m;i++)
{
scanf("%d",&a);
if(a!=0) num[cnt++]=a;
}
//int m=cnt;
int zhuang=1<<cnt;
for(i=1;i<zhuang;i++) //状态
{
int tem=i;
for(j=0;j<cnt;j++) //状态取数情况
{
visi[j]=tem&1;
tem=tem>>1;
}
solve();
}
printf("%I64d\n",sum);
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。