ZOJ 3233 Lucky Number（容斥原理）

Lucky Number
Time Limit: 5 Seconds     
Memory Limit: 32768 KB

Watashi loves M mm very much. One day, M mm gives Watashi a chance to choose a number between
low and high, and if the choosen number is lucky, M mm will marry him.

M mm has 2 sequences, the first one is BLN (Basic Lucky Numbers), and the second one is BUN (Basic Unlucky Numbers). She says that a number is lucky if it's divisible by at least one number from BLN and not divisible by at least one number from BUN.

Obviously, Watashi doesn't know the numbers in these 2 sequences, and he asks M mm that how many lucky number are there in [low,
high]?

Please help M mm calculate it, meanwhile, tell Watashi what is the probability that M mm marries him.

Input

The first line of each test case contains the numbers NBLN (1 <=
NBLN <= 15), NBUN (1 <= NBUN <= 500),
low, high (1 <= low <= high <= 1018).

The second and third line contain NBLN and NBUN integers, respectively. Each integer in sequences BLN and BUN is from interval [1, 32767].

The last test case is followed by four zero.

The input will contain no more than 50 test cases.

Output

For each test case output one number, the number of lucky number between low and
high.

Sample Input

2 1 70 81
2 3
5
0 0 0 0

Sample Output

5

Hint

The lucky numbers in the sample are 72, 74, 76, 78, 81. 

题意：询问[low,high]中，A集合存在一个数是它的约数，且B集合至少存在一个数不是它的约数的个数。

题解：容斥原理。假设B集合中的所有数都是约数，则满足答案的[low,high]之间的数一定可以整除B中所有数的lcm->bm

      这样我们只需要在用容斥原理求[low,high]满足A集合条件的x的时候，减去[low,high]能整除lcm(x,bm)的个数。

      求满足[low,high]中A存在存在一个数是他的约数，把每个数当做一个因子，再容斥。

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <bitset>
#include <vector>

using namespace std;
typedef long long ll;

const int N = 16;
const ll inf=1e18+10;

ll l,r;
int n,m;
int a
,b
;
ll bm;
vector<int>vec;

ll gcd(ll a,ll b) {
return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b) {
if(a==-1||b==-1) return -1;
ll gd=gcd(a,b);
if(inf/b<a/gd) return -1;
return a/gd*b;
}

void init(int x) {
vec.clear();
for(int i=2; i*i<=x; i++) {
if(x%i==0) {
vec.push_back(i);
while(x%i==0)x/=i;
}
}
if(x>1)vec.push_back(x);
}

void solve() {
ll sum=0;
for(int i=1; i<(1<<n); i++) {
ll lc=1;
int cnt=0;
for(int j=0; j<n; j++) {
if(i&(1<<j)) {
lc=lcm(lc,a[j]);
cnt++;
}
}
if(lc==-1)lc=inf;
ll tmp=lcm(bm,lc);
if(tmp==-1) tmp=inf;
ll num=(r/lc-r/tmp)-(l/lc-l/tmp);
if(cnt&1)sum+=num;
else sum-=num;
}
printf("%lld\n",sum);
}

int main() {
//freopen("test.in","r",stdin);
while(~scanf("%d%d%lld%lld",&n,&m,&l,&r)) {
if(n==0&&m==0&&l==0&&r==0)break;
l--;
for(int i=0; i<n; i++)scanf("%d",&a[i]);
bm=1;
for(int i=0; i<m; i++) {
ll x;
scanf("%lld",&x);
bm=lcm(bm,x);
}
solve();
}
return 0;
}