how many integers can you find

HDU-1796,比较基础的容斥原理,求1 ~ n - 1内有多少数能被 M 里面数整除

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

##分析:由题意得只需求M内加上奇数数被n - 1整除,减去偶数数被n - 1整数即可,DFS跑一发,真可啪。

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. **/


#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;

#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define PR(x) printf("%dn", x)
#define PRR(a, b) printf("%d%dn", a, b);
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;

LL qkm(LL a, LL b, LL c){
LL res = 1;

for(a %= c; b; b >>= 1, a = a * a % c) if(b & 1) res = res * a % c;
return res;
}

inline int max(int a, int b){return a > b ? a : b;}
inline int min(int a, int b){return a > b ? b : a;}
struct Node{
string name;
}people[MAX];

long long gcd(long long a,long long b){
return b==0?a:gcd(b,a%b);
}

int cmp(const Node& a, const Node& b){
return a.name.size() < b.name.size();
}

/***********************************************************************************************************************************/
int n, m, len, temp;
LL a[22], ans;

void DFS(int index, LL lcm, int id){ //奇数加,偶数减,最后判断下标即可
lcm = a[index] / __gcd(a[index], lcm) * lcm;

//容斥原理,小于n,故不包含n
if(id & 1) ans += (n - 1) / lcm;
else ans -= (n - 1) / lcm;
//下一个数,lcm,id + 1;
FFI(index + 1, len) DFS(i, lcm, id + 1);
}

int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie(0);

while(cin>>n>>m){
len = ans = 0;
while(m--){
cin>>temp;
if(temp) a[len++] = temp;
}

FFI(0, len) DFS(i, a[i], 1);
cout<<ans<<endl;
}

return EXIT_SUCCESS;
}