HDU How many integers can you find 容斥

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4249    Accepted Submission(s): 1211

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

题意：给n个数字，最大不会超过20的非负数，0忽略它可以。给你一个数字M，

问1-M-1中，有多少个数字能被这n数字中任何一个整除（只要满足其中一个能整除就行）。统计个数输出。

 

思路：容斥，简单容斥。一开始做zoj的一道题，果断数据太水，方法是不对的也能ac。

原来的思路是这样的，对n个数字，筛选掉ai倍数的数字，然后就容斥，但是明显这样的数据有问题

4 6，  这样容斥后得到的结果是4 6 -24，不对的，应该是4 6 -12，所以应该是 4 6    -（4*6）/gcd（4，6）

略坑略坑。

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

bool Hash[22];
int f[22],len,qlen;
__int64 Q[5002];

int gcd(int a,int b)
{
    if(a<0)a=-a;
    if(b<0)b=-b;
    if(b==0)return a;
    int r;
    while(b)
    {
        r=a%b;
        a=b;
        b=r;
    }
    return a;
}
void solve(__int64 m)
{
    qlen = 0;
    Q[0]=-1;
    for(int i=1;i<=len;i++)
    {
        int k=qlen;
        for(int j=0;j<=k;j++)
        Q[++qlen]=-1*(Q[j]*f[i]/gcd(Q[j],f[i]));
    }
    __int64 sum = 0;
    for(int i=1;i<=qlen;i++)
    sum = sum+m/Q[i];
    printf("%I64d
",sum);
}
int main()
{
    int m,x;
    __int64 n;
    while(scanf("%I64d%d",&n,&m)>0)
    {
        n=n-1;
        memset(Hash,false,sizeof(Hash));
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&x);
            Hash[x]=true;
        }
        for(int i=1;i<=20;i++)
        {
            if(Hash[i]==true)
            for(int j=i+i;j<=20;j=j+i)
            if(Hash[j]==true) Hash[j]=false;
        }
        len = 0;
        for(int i=1;i<=20;i++)if(Hash[i]==true) f[++len]=i;
        solve(n);
    }
    return 0;
}