How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1219 Accepted Submission(s): 282
Problem 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
Author
wangye
Source
Recommend
wangye
具体做法是去枚举集合，总共会有1<<m种，这里运用到了二进制位运算。
比起容斥原理中的DFS，我更喜欢这种计算方法。
#include<stdio.h>
__int64 a[15];
__int64 gcd(__int64 a,__int64 b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
__int64 lcm(__int64 a,__int64 b)
{
__int64 g=gcd(a,b);
return a*b/g;
}
int main()
{
__int64 n,m,i,j,k,s,ans,num;
while(scanf("%I64d%I64d",&n,&m)!=EOF)
{
num=0;
for(i=0;i<m;i++)
{
scanf("%I64d",&a[i]);
if(a[i]>0)
a[num++]=a[i];
}
ans=0,m=num;
for(i=1;i<(1<<m);i++)
{
s=1;
k=0;
for(j=0;j<m;j++)
{
if((i>>j)&1)
{
s=lcm(s,a[j]);
k++;
}
}
if(k&1)
ans=ans+(n-1)/s;
else
ans=ans-(n-1)/s;
}
printf("%I64d\n",ans);
}
return 0;
}