How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7439    Accepted Submission(s): 2200


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）
 
题意：给你m个数 可能含有0   问有多少小于n的正数能整除这个m个数中的某一个
题解：特判除零  容斥原理   这m个数 组合时不能直接相乘 应当取最小公倍数
 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 #include<bits/stdc++.h>
 8 #include<map>
 9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define  A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 typedef __int64 ll;
29 const ll MOD=1000000007;
30 const int INF=1000000010;
31 const ll MAX=1ll<<55;
32 const double eps=1e-8;
33 const double inf=~0u>>1;
34 const double pi=acos(-1.0);
35 typedef double db;
36 typedef unsigned int uint;
37 typedef unsigned long long ull;
38 ll gcd(ll aa,ll bb)
39 {
40     if(bb==0)
41         return aa;
42     else
43         return gcd(bb,aa%bb);
44 }
45 ll lcm(ll aa,ll bb)
46 {
47     return aa*bb/gcd(aa,bb);
48 }
49 ll n,m;
50 ll que[15];
51 ll a[10000];
52 ll ggg;
53 ll coun;
54 ll slove(ll gg)
55 {
56     ll t=0,sum=0;
57     a[t++]=-1;
58     for(ll i=0;i<coun;i++)
59     {
60 
61         ll k=t;
62         for(ll j=0;j<k;j++)
63          {
64              a[t++]=que[i]*a[j]*(-1);
65              if(a[t-1]>0)
66                  a[t-1]=lcm(que[i],-a[j]);
67             else
68                  a[t-1]=0-lcm(que[i],a[j]);
69          }
70     }
71     for(ll i=1;i<t;i++)
72             sum=sum+(gg/a[i]);
73     return sum;
74 }
75 int main()
76 {
77     while(scanf("%I64d %I64d",&n,&m)!=EOF)
78     {
79         memset(que,0,sizeof(que));
80         memset(a,0,sizeof(a));
81         coun=0;
82         for(ll i=0;i<m;i++)
83            {
84                scanf("%I64d",&ggg);
85                if(ggg!=0)
86                 que[coun++]=ggg;
87            }
88         printf("%I64d\n",slove(n-1));
89     }
90      return 0;
91 }