HDU 1796 How many integers can you find（简单容斥原理）

题目大意：有一个序列，大小为m，里面有m个不超过20的非负数，各不相同。要求在1-n中有多少个能被m个数中任意一个数整除。

题目思路：简单的容斥原理应用。就不说了直接上代码。

有两种方法，一种是DFS，一种是直接位元素枚举暴力（study from zhixiaoli)

DFS：（速度较快）

#include<iostream>
#include<algorithm>
using namespace std;
long long a[15];
int n, m;
int gcd(int a, int b)
{
if (b == 0)
return 0;
int c = a%b;
while (c)
{
a = b;
b = c;
c = a%b;
}
return b;
}
long long lcm(long long a, long long b)
{
long long c = gcd(a, b);
if (c == 0)
return 0;
return a*b / c;
}
int gao(int x)
{
if (x == 0)
return 0;
return n / x;
}
int dfs(int i, bool flag, long long LCM)
{
long long ans = 0;
long long mid = lcm(max(LCM,a[i]), min(LCM, a[i]));
if (flag) ans += gao(mid);
else ans -= gao(mid);
for (int j = i + 1; j <= m; j++)
ans += dfs(j, !flag, mid);
return ans;
}
int main()
{
while (cin >> n >> m)
{
n--;
for (int i = 1; i <= m; i++)
cin >> a[i];
long long num = 0;
for (int i = 1; i <= m; i++)
num += dfs(i, 1, 1);
cout << num << endl;
}
}

位元素枚举：（速度较慢）

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
int n, m;
ll a[15];
ll GCD(ll x, ll y)
{
return y ? GCD(y, x%y) : x;
}
ll LCM(ll x, ll y)
{
return x / GCD(x, y)*y;
}
ll gao(int l, int&cnt)
{
ll res = 1;
for (int i = cnt = 0; i < m; i++)
{
if (l&(1 << i))
{
cnt++;
res = LCM(res, a[i]);
}
}
return res;
}
int main()
{
while (~scanf("%d%d", &n, &m))
{
n--;
int k=m;
for (int i = 0; i < m; i++)
{
scanf("%lld", &a[i]);
if (!a[i]) k = i;
}
if (k != m)
{
m--;
swap(a[m], a[k]);
}
ll ans = 0;
for (int i = 1; i < (1 << m); i++)
{
ll temp = gao(i, k);
if (k & 1)
ans += n / temp;
else
ans -= n / temp;
}
cout << ans << endl;
}
}