UVa 10325 The Lottery 【容斥】【入门】

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本文解析了UVa10325彩票问题，通过容斥原理求解1-n中不能被给定的m个数整除的数的数量。提供了完整的代码实现，并解释了关键步骤。

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题目链接：UVa 10325 The Lottery

Description：
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 fromthis 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

题目大意：
有m个数，输出1-n中不能被这m个数中任意的数整除的个数。（这m个数中保证不会出现1）

解题思路：
因为n特别大，而m很小，所以直接暴力遍历n是不合理的，这时要考虑根据容斥原理来做。
即总个数-能被其中的一个数整除的数的个数+能被其中两个数都整除的数的个数-能被其中三个数都整除的数的个数……

Mycode：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX = 10005;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;

LL n;
int m;
LL a[20];
LL gcd(LL a, LL b)
{
    return b == 0 ? a : gcd(b, a % b);
}
LL lcm(LL a, LL b)
{
    return a / gcd(a, b) * b;
}
int main()
{
    while(~scanf("%lld%d",&n,&m))
    {
        memset(a, 0, sizeof(a));
        for(int i = 0; i < m; ++i)
            scanf("%lld",&a[i]);

        LL ans = 0;
        for(int i = 1; i < (1 << m); ++i)
        {
            LL mul = 1;
            int bits = 0;
            for(int j = 0; j < m; ++j)
            {
                if(i & (1 << j))
                {
                    bits++;
                    mul = lcm(mul, a[j]);
                }
            }
            if(bits & 1)
                ans += n / mul;
            else
                ans -= n / mul;
        }

        printf("%lld\n", n - ans);
    }
    return 0;
}

【说明】：
1.一个集合的非空子集有2^n个，所以第一层循环遍历了2^n次(也可以遍历2^n-1次，这时就是这个集合本身不被计算在内)
2.如何选取每个子集呢？我们用2^k(1≤k≤n)分别与第一层的i进行与运算(按二进制位考虑，1代表取这个集合，0代表不取这个集合)，这样遍历一遍就能取到所有情况了。
3.bits代表了取了几个集合，然后奇加偶减。
4.1-n中，能被x整除的数的个数为n/x取整。