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UVA10325--- The Lottery (容斥)

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UVA10325--- The Lottery (容斥)

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.
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.
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.
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

容斥原理的簡單應用

/*************************************************************************
    > File Name: uva10325.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月27日 星期三 20時44分29秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair  PLL;

LL num[20];

LL gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}

int main() {
    LL n;
    int m;
    while (cin >> n >> m) {
        for (int i = 0; i < m; ++i) {
            cin >> num[i];
        }
        int ans = 0;
        for (int i = 1; i < (1 << m); ++i) {
            int bits = 0;
            LL cur = 1;
            for (int j = 0; j < m; ++j) {
                if (i & (1 << j)) {
                    ++bits;
                    cur = cur / gcd(num[j], cur) * num[j];
                }
            }
            if (bits & 1) {
                ans += n / cur;
            }
            else {
                ans -= n / cur;
            }
        }
        cout << n - ans << endl;
    }
    return 0;
}