How many integers can you find (HDU 1796)

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

Total Submission(s): 7520    Accepted Submission(s): 2228

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 22 3

 

Sample Output

7

Source

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

题意：有一个数n，和一个拥有m个元素的集合，求在小于n的正整数中能被集合m中任意元素整除的整数个数。

题解：容斥原理，先求出能被一个元素整除的整数个数，再减去能被两个元素整除的整数个数，再加上能被三个元素整除的整数个数，以此类推。

注意：此题在容斥是，计算的是k个元素最小公倍数，而不是乘积。

代码：

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566

#include<iostream>#include<cstdio>using namespace std;long long s;long long m;int n;int count_=1;int a[25];int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int lcm(int a,int b){    return a/gcd(a,b)*b;}void dfs(int k,long long ans,int cou,int j){    if(cou==k)    {        s+=(m-1)/ans;        return ;    }    for(int i=j; i<=n; i++)    {        int aa=ans;        ans=lcm(ans,a[i]);        cou++;        dfs(k,ans,cou,i+1);        cou--;        ans=aa;    }    return ;}int main(void){    while(~scanf("%lld%d",&m,&n))    {        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        for(int i=1;i<=n;i++)        {            if(a[i]==0)            {                for(int j=i;j<=n;j++)                    a[j]=a[j+1];                n--;            }        }        long long sum=0;        for(int i=1; i<=n; i++)            sum+=((m-1)/a[i]);        for(int i=2; i<=n; i++)        {            s=0;            dfs(i,1,0,1);            if(i%2==0)sum+=(-1)*s;            else sum+=s;        }        printf("%lld\n",sum);        count_=1;    }    return 0;}