HDU 1796 容斥原理

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

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<map>
 #include<set>
 #include<cmath>
 #include<queue>
 #include<bitset>
 #include<math.h>
 #include<vector>
 #include<string>
 #include<stdio.h>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #pragma comment(linker, "/STACK:102400000,102400000")
 using namespace std;
 #define  A first
 #define B second
 const int mod=;
 const int MOD1=;
 const int MOD2=;
 const double EPS=0.00000001;
 typedef __int64 ll;
 const ll MOD=;
 const int INF=;
 const ll MAX=1ll<<;
 const double eps=1e-;
 const double inf=~0u>>;
 const double pi=acos(-1.0);
 typedef double db;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 ll gcd(ll aa,ll bb)
 {
     if(bb==)
         return aa;
     else
         return gcd(bb,aa%bb);
 }
 ll lcm(ll aa,ll bb)
 {
     return aa*bb/gcd(aa,bb);
 }
 ll n,m;
 ll que[];
 ll a[];
 ll ggg;
 ll coun;
 ll slove(ll gg)
 {
     ll t=,sum=;
     a[t++]=-;
     for(ll i=;i<coun;i++)
     {

         ll k=t;
         for(ll j=;j<k;j++)
          {
              a[t++]=que[i]*a[j]*(-);
              if(a[t-]>)
                  a[t-]=lcm(que[i],-a[j]);
             else
                  a[t-]=-lcm(que[i],a[j]);
          }
     }
     for(ll i=;i<t;i++)
             sum=sum+(gg/a[i]);
     return sum;
 }
 int main()
 {
     while(scanf("%I64d %I64d",&n,&m)!=EOF)
     {
         memset(que,,sizeof(que));
         memset(a,,sizeof(a));
         coun=;
         for(ll i=;i<m;i++)
            {
                scanf("%I64d",&ggg);
                if(ggg!=)
                 que[coun++]=ggg;
            }
         printf("%I64d\n",slove(n-));
     }
      return ;
 }