HDU

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

题意：给出n和m，然后给出一个m个数的集合，让求小于n的数（1~n-1）里面可以被集合中至少一个数整除的数的个数，

这是一个容斥定理的简单的应用题，做这个题思路还是很清晰的，首先第一个坑就是，a[i]可能为0，而0是不能做除数的，所以要去掉集合中的0，那么接下来用容斥原理的思路就是，找出1~n-1中可以被每个a[i]整除的数的个数，减去可以被集合中任意两个数整除的数的个数，即可以整除任意两个数最小公倍数的数的个数，加上可以被集合中任意三个数整除的数的个数。。。。。。

关于二进制运算原理参考：HDU - 4135 - Co-prime - （容斥原理，分解质因数）

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 15ll n,m,num;ll a[M];ll gcd(ll a,ll b){    return b==0?a:gcd(b,a%b);}ll lcm(ll a,ll b){    return a/gcd(a,b)*b;}ll solve(){    ll sum=0;    for(ll msk=1;msk<(1ll<<num);msk++) //枚举集合中所有数的所有组合情况    {        bool flag=true;        ll mult=1,bits=0;        for(ll i=0;i<num;i++)          //判断哪些a[i]在当前排列中        {            if(msk&(1ll<<i))           //在排列中            {                ll temp=lcm(a[i],mult); //求其最小公倍数                if(temp<n)             //必须是小于n的数                {                    ++bits;            //记录个数                    mult=temp;                }                else{                    flag=false;                    break;                }            }        }        if(flag==false)               //如果当前组合已经大于n是不符合题意的            continue;        ll cur=(n-1)/mult;            //可以被mult整除的数的个数        if(bits&1)                    //奇加偶减            sum+=cur;        else            sum-=cur;    }    return sum;}int main(){    int i;    while(scanf("%lld%lld",&n,&m)!=EOF)    {        ll t;        num=0;        for(i=0;i<m;i++)        {            scanf("%lld",&t);            if(t!=0)            {                a[num++]=t;            }        }        sort(a,a+num);        printf("%lld\n",solve());    }    return 0;}

看到dfs的解法：http://blog.csdn.net/wyt734933289/article/details/51292489，这里dfs要比简单容斥快

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;int n, m;int a[25];ll ans, lcm;ll gcd(ll a, ll b){    return b == 0 ? a : gcd(b, a%b);}void dfs(int cur, ll val, int flag){    for(int i = cur; i <= m; i++)    {        if(a[i])        {            lcm = a[i] / gcd(val, a[i]) *val;            if(lcm > n) continue;//当lcm > n,就没有必要继续下去，因为再怎么加 n/lcm也是0，ans不会有任何变化            ans += flag * ( n / lcm );            dfs(i + 1, lcm, -flag);        }    }}int main(){    while(~scanf("%d%d", &n, &m))    {        for(int i = 1; i <= m; i++)        {            scanf("%d", &a[i]);        }        n--;        ans = 0;        dfs(1, 1, 1);        printf("%I64d\n", ans);    }    return 0;}