uva 10325 容斥原理

The Sports Association of Bangladesh is in great problem with their latest lottery 'Jodi laiga Jai'. There are so many participants this time that they cannot manage all the numbers. In an urgent meeting they have decided that they will ignore some numbers. But how they will choose those unlucky numbers!! Mr. NondoDulal who is very interested about historic problems proposed a scheme to get free from this problem.

You may be interested to know how he has got this scheme. Recently he has read the Joseph's problem.

The Problem

There are N tickets which are numbered from 1 to N. Mr. Nondo will choose M random numbers and then he will select those numbers which is divisible by at least one of those M numbers. The numbers which are not divisible by any of those M numbers will be considered for the lottery.

As you know each number is divisible by 1. So Mr. Nondo will never select 1 as one of those M numbers. Now given N,M and M random numbers, you have to find out the number of tickets which will be considered for the lottery.

The Input

Each input set starts with two Integers N (10<=N<2^31) and M (1<=M<=15). The next line will contain M positive integers each of which is not greater than N. Input is terminated by EOF.

The Output

Just print in a line out of N tickets how many will be considered for the lottery.

Sample Input

10 2
2 3
20 2
2 4

Sample Output

3
10

先求出不符合的数量 sum ：

举例：m = 5

对 m 个数字进行组合，用二进制枚举所有情况，求每组组合数的最小公倍数。

flag 表示组合数的数量

if(flag%2==0) sum -= N(n,g); //偶数个
else sum+=N(n,g);  //奇数个

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long data[20];
long long gcd(long long a,long long b){
    if(b==0) return a;
    return gcd(b,a%b);
}
long long lcm(long long a,long long b){
    return (long long)a/gcd(a,b)*b;
}
long long N(long long n,long long x){
    return n/x;
}
int main(){
    long long n,m;
    long long sum,g,flag;
    while(scanf("%lld%lld",&n,&m)!=EOF){
        sum = 0;
        for(int i=0;i<m;i++) {
            scanf("%d",&data[i]);
        }
        sum = 0;
        for(int i=1;i<(1<<m);i++) {
            g = 1;
            flag = 0;
            for(int j=0;j<m;j++) {
                if((1<<j)&i) {
                    g = lcm(g,data[j]);
                    if(g>n) break;
                    flag++;
                }
            }
            if(flag%2==0) sum -= N(n,g);
            else sum+=N(n,g);
        }
        printf("%lld\n",n-sum);
    }
}