HDU 1796 How many integers can you find

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5517    Accepted Submission(s): 1580

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
  

 

Author

wangye

 

Source

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

 

Recommend

wangye

 

题目意思是给出数n和一个包含m个数的集合，要求的是小于n，且是集合内任意元素的倍数的数有多少。（有点绕，多读两遍就理解了）。

这个题目要用到的原理是容斥原理，什么是容斥原理呢？大概学过集合或者概率论的都知道吧！

容斥原理：先不考虑重叠的情况，把包含于某内容中的所有对象的数目先计算出来，然后再把计数时重复计算的数目排斥出去，使得计算的结果既无遗漏又无重复。

本题的关键是要如何实现容斥原理，最简单的就是直接暴力，数据比较水能过，还有一种方法就是用dfs优化。

两种方法均可，只是时间复杂度的大小问题。

此题还有一点点小坑，稍不细心的人就可能掉进去（比如我……T_T……），集合中元素可能包含0，要先筛选一遍才能操作……不多说了，直接上代码：

/*
Author:ZXPxx
Memory: 1616 KB		Time: 202 MS
Language: C++		Result: Accepted
*/

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 5e5 + 5;
const int inf = 0x3f3f3f3f;

int a[130],n,m,ans,p,cur;
int gcd(int a,int b) {
    return !b ? a:gcd(b,a%b);
}
int lcm(int a,int b) {
    return a/gcd(a,b)*b;
}
void dfs(int i,int flag,int k)
{
    k=lcm(a[i],k);
    ans+=n/(k*flag);
    for(int j=i+1;j<cur;j++)
        dfs(j,-flag,k);
}
int main() {
    while(~scanf("%d %d",&n,&m)) {
        cur=0;
        for(int i=0; i<m; i++) {
            scanf("%d",&p);
            if(p)
                a[cur++]=p;
        }
        ans=0;n--;
        for(int i=0; i < cur; i++) {
            dfs(i,1,a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}

爆搜的代码也来一发。。。

/*
Author:ZXPxx
Memory: 1648 KB		Time: 998 MS
Language: C++		Result: Accepted
*/

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 5e5 + 5;
const int inf = 0x3f3f3f3f;

int a[130],m,n,p,cur,ans;
int gcd(int a,int b) {
    return !b ? a:gcd(b,a%b);
}
int lcm(int a,int b) {
    return a/gcd(a,b)*b;
}
int main() {
    while(~scanf("%d %d",&n,&m)) {
        n--;    cur=0;     ans=0;
        for(int i=0; i<m; i++) {
            scanf("%d",&p);
            if(p)  a[cur++]=p;
        }
        for(int i= 1;i < pow(2,cur); i++) {
            int cnt=0,flag=0,k=1;
            for(int j = 0; j < cur; j++) {
                if(i>>j & 1) {
                    k=lcm(k,a[j]);
                    if(k>n) {
                        flag=1;
                        break;
                    }
                    cnt++;
                }
            }
            if(flag) continue;
            if(cnt%2)
                ans+=n/k;
            else
                ans-=n/k;
        }
        printf("%d\n",ans);
    }
    return 0;
}