HDU 1796How many integers can you find（容斥原理）

How many integers can you find

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1796

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

分析：这几天练容斥有感觉，知道是容斥，但是却有问题，容斥是 互质的数，然后对于2,4这样的数就不会做了，太肤浅了，直接求最小公倍数啊，对啊，互质的相乘就是因为最小公倍数就是乘积啊=_=，弱！

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 int num[], n, m, a[];
 LL res;
 LL gcd(LL a, LL b)
 {
     if (a == )
         return b;
     return gcd(b % a, a);
 }
 void dfs(int cur, int snum, int cnt)
 {
     if (snum == cnt)
     {
         int temp = n;
         int mult = ;
         for (int i = ; i < snum; i++)
             mult = mult / gcd(mult, a[i]) * a[i];  // 防爆
         if (mult == )
             return;
         if (temp % mult == )
             res += temp / mult - ;
         else
             res += temp / mult;
         return;
     }
     for (int i = cur; i < m; i++)
     {
         a[snum] = num[i];
         dfs(i + , snum + , cnt);
     }
 }
 int main()
 {
     int tm;
     while (scanf("%d%d", &n, &tm) != EOF)
     {
         m = ;
         for (int i = ; i < tm; i++)
         {
             int temp;
             scanf("%d", &temp); // 去0
             if (temp)
                 num[m++] = temp;
         }
         LL sum = ;
         for (int i = ; i <= m; i++)
         {
             res = ;
             dfs(, , i);
             if (i & )
                 sum += res;
             else
                 sum -= res;
         }
         printf("%I64d\n", sum);
     }
     return ;
 }