HDU 1796 - How many integers can you find (容斥原理)
题意
输出小于N的,能整除一个集合里的数的个数。
思路
最基本的容斥原理。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-8;
const int MAXN = 60;
const int MOD = 1000007;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int ans, n, pivot;
vector<int> num;
int LCM(int a, int b)
{
return a / __gcd(a, b) * b;
}
int DFS(int curPos, int curPro, int curCnt)
{
for (int i = curPos; i < SZ(num); i++)
{
int tmp = (pivot - 1) / LCM(curPro, num[i]);
if (curCnt & 1) ans += tmp;
else ans -= tmp;
DFS(i + 1, LCM(curPro, num[i]), curCnt + 1);
}
}
int main()
{
//ROP;
int i, j;
while (cin >> pivot >> n)
{
num.clear();
ans = 0;
for (i = 0; i < n; i++)
{
int tmp;
scanf("%d", &tmp);
if (tmp) num.PB(tmp);
}
DFS(0, 1, 1);
printf("%d\n", ans);
}
return 0;
}