How many integers can you find
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
分析：这几天练容斥有感觉，知道是容斥，但是却有问题，容斥是 互质的数，然后对于2,4这样的数就不会做了，太肤浅了，直接求最小公倍数啊，对啊，互质的相乘就是因为最小公倍数就是乘积啊=_=，弱！
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
int num[], n, m, a[];
LL res;
LL gcd(LL a, LL b)
{
if (a == )
return b;
return gcd(b % a, a);
}
void dfs(int cur, int snum, int cnt)
{
if (snum == cnt)
{
int temp = n;
int mult = ;
for (int i = ; i < snum; i++)
mult = mult / gcd(mult, a[i]) * a[i]; // 防爆
if (mult == )
return;
if (temp % mult == )
res += temp / mult - ;
else
res += temp / mult;
return;
}
for (int i = cur; i < m; i++)
{
a[snum] = num[i];
dfs(i + , snum + , cnt);
}
}
int main()
{
int tm;
while (scanf("%d%d", &n, &tm) != EOF)
{
m = ;
for (int i = ; i < tm; i++)
{
int temp;
scanf("%d", &temp); // 去0
if (temp)
num[m++] = temp;
}
LL sum = ;
for (int i = ; i <= m; i++)
{
res = ;
dfs(, , i);
if (i & )
sum += res;
else
sum -= res;
}
printf("%I64d\n", sum);
}
return ;
}