hdu 1796 How many integers can you find 容斥定理

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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）

思路：最简单的容斥，注意下可能输入0；奇加偶减

　　　比如12 2 

              2   3

　　　　　ans=11/2+11/3-11/6;

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll a[110];
ll ji;
ll ans,x,y;
ll gcd(ll x,ll y)
{
    return y==0?x:gcd(y,x%y);
}
void dfs(ll lcm,ll pos,ll step)
{
    if(lcm>x)
    return;
    if(pos==ji)
    {
        if(step==0)
            return;
        if(step&1)
        ans+=x/lcm;
        else
        ans-=x/lcm;
        return;
    }
    dfs(lcm,pos+1,step);
    dfs(lcm/gcd(a[pos],lcm)*a[pos],pos+1,step+1);
}
int main()
{
    ll z,i,t;
    while(~scanf("%I64d%I64d",&x,&y))
    {
        x--;
        ji=0;
        for(i=0;i<y;i++)
        {
            scanf("%I64d",&z);
            if(z==0)continue;
            a[ji++]=z;
        }
        ans=0;
        dfs(1,0,0);
        printf("%I64d
",ans);
    }
    return 0;
}