Hdu1796 How many integers can you find

How many integers can you find

Time Limit : 12000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 95   Accepted Submission(s) : 30

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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 22 3

Sample Output

7

Author

wangye

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

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题目的意思是给出一个n和m个数求小于n的有多少个是m个数任意个的倍数

求出，每个数和他们任意的倍数的个数利用容斥原理解决 在求个数时可以用DFS

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <cmath>using namespace std;#define LL long longconst int inf=0x7fffffff;LL a[100];struct node{    LL num;    int cnt;}ans[100005];int tot,m;LL n;LL gcd(LL a,LL b){    return b==0?a:gcd(b,a%b);}void dfs(int pos,LL lcm,int cnt){    if(pos>=m)    {        if(cnt==0)            return;        ans[tot].num=(n-1)/lcm;        ans[tot++].cnt=cnt;        return;    }    if(a[pos]==0)        dfs(pos+1,lcm,cnt);    else    {        dfs(pos+1,lcm*a[pos]/gcd(lcm,a[pos]),cnt+1);        dfs(pos+1,lcm,cnt);    }}int main(){    while(~scanf("%lld%d",&n,&m))    {        for(int i=0; i<m; i++)        {             scanf("%lld",&a[i]);        }        tot=0;        dfs(0,1,0);        LL ass=0;        for(int i=0; i<tot; i++)        {            if(ans[i].cnt%2)                ass+=ans[i].num;            else                ass-=ans[i].num;        }        printf("%lld\n",ass);    }    return 0;}