hdu 1796 How many integers can you find【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): 4674    Accepted Submission(s): 1340

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
   
   
   
   

题目翻译：给出N和M，接下来给出一个M个元素的集合，求1-（N-1）中，有多少个数字是集合中任意元素的倍数！最后只写个数！

解题思路：还是利用容斥原理来求，只不过这次给你的集合貌似就是以前所求的p数组元素，但是他和p数组元素的要求又有所不同，因为以前所求的p数组元素只有素数，且任意元素都不相同，但是这一次的却是很混杂，只说是给的非负数，因此我们在求的时候首先要求掉一些无关元素，比如：0和大于N的数字，这一步输入的时候处理！

再其次，容斥原理减去重叠元素的时候，按照以前的，假如素因子分别是A和B，则我们要减去N/(A*B),但是对于本题，由于A,B可能存在公约数，我们却要算的是N除以lcm（A,B），lcm是最小公倍数，具体算法如下：

#include <stdio.h>
#include <algorithm>
using namespace std;
int p[12],top,ans,n;
int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
int nop(int m){
    int i,j,sum=0,flag,num;
    for(i=1;i<1<<top;i++){
        flag=0;
        num=1;
        for(j=0;j<top;j++)
            if(i&(1<<j))
                flag++,num=lcm(num,p[j]);
        if(flag&1) sum+=m/num;
        else       sum-=m/num;
    }
    return sum;
}
int main(){
    while(~scanf("%d%d",&n,&top)){
        for(int i=0;i<top;i++){
            scanf("%d",&p[i]);
            if(p[i]<1||p[i]>=n)
                i--,top--;
        }
        printf("%d\n",nop(n-1));
    }
    return 0;
}