UVA10325 The Lottery（容斥原理）

最新推荐文章于 2022-07-16 21:11:29 发布

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本文提供了一种利用容斥原理解决特定数值范围内的问题，即计算给定区间内有多少个数不是指定一组数的倍数。通过深入探讨容斥原理的应用，该文详细阐述了算法实现过程，并提供了相应的代码示例。

题意：

给n,m,和m个数（k1~km）。求1~n中有多少个数不是(k1~km)中任意一数的倍数。

题解：

容斥模板题。反面考虑，a的倍数有n/a个；既是a,也是b的倍数，即lcm(a,b)的倍数有n/lcm(a,b)个。是a,b,c的倍数，即lcm(a,b,c)的倍数有n/lcm(a,b,c)个。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
LL a[20],m;
LL n,ans;
 
LL lcm(LL a,LL b)
{
    return a/__gcd(a,b)*b;
}
 
void dfs(int c,int cur,int i,LL ans1)      //dfs(c,1,i,0,1);
{
    if(cur==c+1)
    {
        if(c&1)
            ans-=n/ans1;
        else
            ans+=n/ans1;
        return;
    }
    for(;i<m;i++)
    {
        dfs(c,cur+1,i+1,lcm(ans1,a[i]));
    }
}
 
int main()
{
    while(cin>>n>>m)
    {
        for(int i=0;i<m;i++)
            scanf("%lld",&a[i]);
        ans=n;
        for(int c=1;c<=m;c++)
            dfs(c,1,0,1);
        printf("%lld\n",ans);
    }
}

The Lottery

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

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