hdu 1796(容斥原理+状态压缩)

Posted 2020-07-16 AC菜鸟机

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6630    Accepted Submission(s): 1913

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

 

题意:在 1-(n-1) 中能够被输入的数字整除的数字的数量。

思路:容斥原理+枚举状态,碰到奇数加上(n-1)/lcm(a,b,c..) 碰到偶数减(n-1)/lcm(a,b,c...) 注意0不能取，发现本人一直不是很会用深搜，所以还是用状压了 = =

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
int main()
{
    int n,m,a[15];
    while(scanf("%d%d",&n,&m)!=EOF){
        int id = 0,num;
        for(int i=0;i<m;i++){
            scanf("%d",&num);
            if(num!=0) a[id++] = num;
        }
        int ans = 0;
        for(int i=1;i<(1<<id);i++){
            int l=1,cnt=0;
            for(int j=0;j<id;j++){
                if((i>>j)&1){
                    cnt++;
                    l = lcm(l,a[j]);
                }
            }
            if(cnt&1){
                ans+=(n-1)/l; ///不包括自身所以n-1
            }else{
                ans-=(n-1)/l;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}