[容斥原理] zoj 2836 Number Puzzle

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题目链接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1836

Given a list of integers (A₁, A₂, ..., A_n), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A₁, A₂, ..., A_n(1 <= A_i <= 10, for i = 1, 2, ..., N).

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2
2 3 7
3 6
2 3 7

Sample Output

1
4

题目意思：

给n个数，求不超过m的能被这n个数中任意一个整除的数的个数。

解题思路：

简单容斥原理

先求出能被一个数整除的数的个数，再减去能被两个数整除的，再加上能被三个数整除的.....

可以用dfs模拟求。

注意能被多个数整除的，要求出他们的最小公倍数。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 11
int n,m;

int vis[Maxn],va[Maxn],cnt;
LL ans;

LL gcd(LL a,LL b)
{
    if(a%b==0)
        return b;
    return gcd(b,a%b);
}
LL lcm(LL a,LL b)
{
    return a/gcd(a,b)*b;
}
void dfs(LL le,int cur,int num)
{
    if(num>cnt||le>m)
        return ;
    for(int i=cur;i<=cnt;i++)
    {
        LL temp=lcm(le,va[i]);
        if(num&1)
            ans+=m/temp;
        else
            ans-=m/temp;
        dfs(temp,i+1,num+1);
    }
}

int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d%d",&n,&m))
   {
       cnt=0;
       memset(vis,0,sizeof(vis));

       for(int i=1;i<=n;i++)
       {
           int a;

           scanf("%d",&a);
           if(!vis[a])
           {
               vis[a]=1;
               va[++cnt]=a;
           }
       }
       ans=0;
       for(int i=1;i<=cnt;i++)
       {
           ans+=m/va[i];
           dfs(va[i],i+1,2);
       }
       printf("%lld\n",ans);
   }
   return 0;
}

[容斥原理] zoj 2836 Number Puzzle,布布扣,bubuko.com

[容斥原理] zoj 2836 Number Puzzle

标签：style   blog   class   code   tar   ext   

原文地址：http://blog.csdn.net/cc_again/article/details/25080723