首页 购物 网址 三丰软件 | 小说 美女秀 图库大全 游戏 笑话 | 下载 开发知识库 新闻 开发 图片素材
 俄罗斯方块 ↓俄罗斯方块↓ TxT小说阅读器 ↓语音阅读,小说下载,古典文学↓ 一键清除垃圾 ↓轻轻一点,清除系统垃圾↓ 图片批量下载器 ↓批量下载图片,美女图库↓ 移动开发 架构设计 编程语言 Web前端 互联网 开发杂谈 系统运维 研发管理 数据库 云计算 Android开发资料
 资讯 业界资讯 软件杂谈 编程开发 网站建设 网络观查 搜索引擎 移动应用 网站运营 网络地图 开发 移动开发 Web前端 架构设计 编程语言 互联网 数据库 系统运维 云计算 开发杂谈
 [编程语言] UVA 10325 The Lottery(容斥）
 UVA 10325 The Lottery(容斥）

# The Lottery

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.

```10 2
2 3
20 2
2 4
```

## Sample Output

```3
10
```
```
Md. Kamruzzaman把出局的人找出来拿n减去出局的人就可以了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int MOD=1e9+7;
typedef long long LL;
vector<int>v;
LL n,m;
LL a;
LL gcd(LL a,LL b)
{
return b?gcd(b,a%b):a;
}
LL lcm(LL a,LL b)
{
return a/gcd(a,b)*b;
}
void solve()
{
LL ans=0,lc;
for(int i=1;i<(1<<m);i++)
{
v.clear();
for(int j=0;j<m;j++)
{
if(i&(1<<j))
v.push_back(a[j]);
}
lc=1;
for(int j=0;j<v.size();j++)
lc=lcm(lc,v[j]);
if(v.size()&1)
ans+=n/lc;
else
ans-=n/lc;
}
cout<<n-ans<<endl;
}
int main()
{
while(cin>>n>>m)
{
REP(i,m)   cin>>a[i];
solve();
}
return 0;
}

```

 此文从网络中自动搜索生成，不代表本网站赞成被搜索网站的内容或立场    查看原文