hdoj 1796 How many integers can you find<容斥原理>

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6741    Accepted Submission(s): 1957


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

2008
“Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

代码：

#include<cstdio>
#include<cstring>
int n,m,s;
int shu[12],mm;
int qu[1000010];
int gcd(int dd,int xx)
{
if (dd%xx)
return gcd(xx,dd%xx);
return xx;
}
int GCD(int fu,int zz)
{
int lp=1,a;
if (fu<0)
{
lp=-1;
fu=-fu;
}
if (fu>zz)
{
a=fu;fu=zz;zz=a;
}
if (zz%fu)
return zz*fu/gcd(fu,zz%fu)*lp;
return zz*lp;
}
int main()
{
while (~scanf("%d%d",&n,&m))
{
bool fafe[10];
memset(fafe,false,sizeof(fafe));
for (int i=0;i<m;i++)
{
scanf("%d",&shu[i]);
}
mm=0;
for (int i=0;i<m;i++)
if (shu[i])
shu[mm++]=shu[i];
int ge=0;
memset(qu,0,sizeof(qu));
qu[ge++]=-1;
for (int i=0;i<mm;i++)
{
int kk=ge;
for (int j=0;j<kk;j++)
qu[ge++]=GCD(qu[j],shu[i])*(-1);
}
s=0;n--;
for (int i=1;i<ge;i++)
{
s+=n/qu[i];
}
printf("%d\n",s);
}
return 0;
}