LuoGu: CF449D Jzzhu and Numbers
CF: D. Jzzhu and Numbers
使得与和为一个数，需要所有数都含有这个数，而不能有多的位。对于一个数，考虑其超集，在超集中选择任意的数，一定与和是这个数的超集，除非不选，则共有 $2 ^ s - 1$ 中方案。考虑容斥，需要排除多的位，再将超集的答案减去即可。
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| #include <cstdio>
using namespace std;
typedef long long LL;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 20, mod = 1e9 + 7;
int n, f[1 << N];
int binpow(int b, int k)
{
int res = 1;
while (k)
{
k & 1 && (res = (LL)res * b % mod);
b = (LL)b * b % mod;
k >>= 1;
}
return res;
}
int main()
{
read(n);
for (int a; n; n--)
read(a), f[a]++;
for (int i = 0; i < N; i++)
for (int j = 0; j < 1 << N; j++)
j >> i & 1 && ((f[j ^ 1 << i] += f[j]) %= mod);
for (int i = 0; i < 1 << N; i++)
f[i] = binpow(2, f[i]) - 1;
for (int i = 0; i < N; i++)
for (int j = 0; j < 1 << N; j++)
j >> i & 1 && ((f[j ^ 1 << i] -= f[j]) %= mod);
write((f[0] + mod) % mod);
return 0;
}
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