How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6710    Accepted Submission(s): 1946

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

题目大意:给定n和一个大小为m的集合,集合元素为非负整数。为1...n内能被集合里任意一个数整除的数字个数。n<=2^31,m<=10

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-;
const int inf =0x7f7f7f7f;
const double pi=acos(-);
const int maxn=; ll gcd(ll a,ll b)
{
if(b==) return a;
else return gcd(b,a%b);
} ll lcm(ll a,ll b)
{
return (a/gcd(a,b))*b;
} int n,m,bit,mm[],tmp,cnt;
ll mult;
void solve(int flag)
{
mult=;bit=;
for(int i=;i<cnt;i++)
if(flag&(<<i))
{mult=lcm(mm[i],mult);bit++;}
} int main()
{
while(~scanf("%d %d",&n,&m))
{
ll ans=;n--;cnt=;
for(int i=;i<m;i++)
{
scanf("%d",&tmp);
if(tmp) mm[cnt++]=tmp;
}
for(int i=;i<(<<cnt);i++)
{
solve(i);
int num=((ll)n)/mult;
if(bit%==) ans+=num;
else ans-=num;
}
printf("%lld\n",ans);
}
return ;
}

分析:很好的一道容斥题,分析:求出在给定区间中能被集合中任意一个数整除的点的个数,分析题目的话
可以发现,先求出区间中所有能被集合中单个数整除的点的个数,求和后,会发现,能同时被两个数整除的点(是这两个数的最小公倍数的倍数)多算了一次,所以就减去能同时被两个数整除点的总个数,然后再加上能同时被三个点减去的点的个数.....(容斥),不过这个题目有个很大的坑点,就是必须要去0,否则不仅会导致re,而且还会直接导致错误,因为在下面这段代码中,如果cnt换成m的话。可以发现0的存在就直接导致了ans的值得变化,所以必须要在读入集合时就直接将0剔除

for(int i=1;i<(1<<cnt);i++)
{
solve(i);
int num=((ll)n)/mult;
if(bit%2==1) ans+=num;
else ans-=num;
}

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