HDU 1796 How many integers can you find （简单容斥）

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传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1796

题意：给你m个正整数，问你从1到n-1有多少个数至少能被这m个数中的一个数整除。

题解：

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s)  scanf("%s",s)#define pi1(a)    printf("%d\n",a)#define pi2(a,b)  printf("%d %d\n",a,b)#define mset(a,b)   memset(a,b,sizeof(a))#define forb(i,a,b)   for(int i=a;i<b;i++)#define ford(i,a,b)   for(int i=a;i<=b;i++)typedef __int64 LL;const int N=40000;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-7;int n,m,cnt;LL a[12];LL gcd(LL a,LL b){    while(b)    {        LL t=a%b;        a=b;        b=t;    }    return a;}//容斥void dfs(int now,int count,LL lcm,LL &ans){    lcm=a[now]/gcd(lcm,a[now])*lcm;    if(count&1) ans+=(n-1)/lcm;    else        ans-=(n-1)/lcm;    for(int i=now+1;i<cnt;i++)        dfs(i,count+1,lcm,ans);}int main(){//    freopen("input.txt","r",stdin);    int i,j;    while(cin>>n>>m)    {        cnt=0;        for(int i=0;i<m;i++)        {            int x;            scanf("%d",&x);            if(x)                a[cnt++]=x;        }        LL ans=0;        for(int i=0;i<cnt;i++)        {            dfs(i,1,a[i],ans);        }        printf("%I64d\n",ans);    }    return 0;}