HDU 1796 容斥原理

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7439    Accepted Submission(s): 2200

Problem Description

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

 

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

 

Output

  For each case, output the number.

 

Sample Input

12 2

2 3

 

Sample Output

7

 

Author

wangye

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

 

题意：给你m个数 可能含有0   问有多少小于n的正数能整除这个m个数中的某一个

题解：特判除零  容斥原理   这m个数 组合时不能直接相乘 应当取最小公倍数

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
#include<bits/stdc++.h>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define  A first
#define B second
const int mod=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-8;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
ll gcd(ll aa,ll bb)
{
    if(bb==0)
        return aa;
    else
        return gcd(bb,aa%bb);
}
ll lcm(ll aa,ll bb)
{
    return aa*bb/gcd(aa,bb);
}
ll n,m;
ll que[15];
ll a[10000];
ll ggg;
ll coun;
ll slove(ll gg)
{
    ll t=0,sum=0;
    a[t++]=-1;
    for(ll i=0;i<coun;i++)
    {

        ll k=t;
        for(ll j=0;j<k;j++)
         {
             a[t++]=que[i]*a[j]*(-1);
             if(a[t-1]>0)
                 a[t-1]=lcm(que[i],-a[j]);
            else
                 a[t-1]=0-lcm(que[i],a[j]);
         }
    }
    for(ll i=1;i<t;i++)
            sum=sum+(gg/a[i]);
    return sum;
}
int main()
{
    while(scanf("%I64d %I64d",&n,&m)!=EOF)
    {
        memset(que,0,sizeof(que));
        memset(a,0,sizeof(a));
        coun=0;
        for(ll i=0;i<m;i++)
           {
               scanf("%I64d",&ggg);
               if(ggg!=0)
                que[coun++]=ggg;
           }
        printf("%I64d
",slove(n-1));
    }
     return 0;
}