How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

题目链接:点击打开链接

给出n, m, n代表1 - n的一个序列, 接下来m个数组成的集合, 问序列中能够整除任一集合中的一个数的个数和为多少.

对读入的m个数进行推断, 非0则赋值到a数组中, 进行dfs, dfs时进行容斥运算, id为奇数则加, 为偶数则减去反复的.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 15;
int n, m, num, ans, a[MAXN];
int gcd(int a, int b)
{
	return b == 0 ? a : gcd(b, a % b);
}
void dfs(int cur, int lcm, int id)
{
	lcm = a[cur] / gcd(a[cur], lcm) * lcm;
	if(id & 1) ans += (n - 1) / lcm;
	else ans -= (n - 1) / lcm;
	for(int i = cur + 1; i < num; ++i)
		dfs(i, lcm, id + 1);
}
int main(int argc, char const *argv[])
{
	while(scanf("%d%d", &n, &m) != EOF) {
		num = ans = 0;
		while(m--) {
			int x;
			scanf("%d", &x);
			if(x != 0) a[num++] = x;
		}
		for(int i = 0; i < num; ++i)
			dfs(i, a[i], 1);
		printf("%d\n", ans);
	}
	return 0;
}