SOJ 3137: Simple Computing _容斥原理

Time Limit: 2000 MS Memory Limit: 65536 K

Description

Given n integers x1 x2 ... xn, you should count how many intergers from 1 to m that can be divided by at least one of them.

Input

The first line is an integer c which shows the number of cases. Each case contains 2 lines. The first line has 2 integers n and m. The second line contains n integers. (0 < n < 11, 0 < m < 2^31, 0 < xi < 100)

Output

For each case print the answer in a single line.

Sample Input

21 1022 203 4

Sample Output

510

Source

The Star of ACMClub 2007 @ HIT

//题意，给出n个数，求1到m中有多少个数(假设这个数是i)满足n个数中至少有一个数能够整除i；

//变形的求1~n中与m互质的数的个数问题！这里将这n个数直接当做是n个素因子，可以产生多种组合，不断筛选出能够整除1~m中得数的个数，容斥！注意的是，假如n中有两个数3和6，而m=18，对于3*6|18，计算值是1，但是12显然也能够被3、6、3*6整除，所以需要求出这k个数的最小公倍数（需要用long long，没注意错了几次）。

#include<stdio.h>
#include<string.h> 
using namespace std;
#define LL long long
#define maxn 15

int a[maxn];

int gcd(int a,int b)
{
	int c;
	while(b)
	{
		c=b;
		b=a%b;
		a=c;
	}	
	return a;
}

int make_ans(int n,int m)
{
	int ans=0,i,j,flag;
	LL tmp;
	for(i=1;i<(1<<n);i++)
	{
		tmp=1,flag=0;
		for(j=0;j<n&&tmp<=m;j++)
			if(i&(1<<j))
				flag++,tmp=(LL)(a[j]/gcd(tmp,a[j]))*tmp;
		if(flag&1)
			ans+=m/tmp;
		else
			ans-=m/tmp;		
	}
	return ans;
}

int main()
{
	int t,n,m,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		printf("%d\n",make_ans(n,m));
	}
	return 0;
}