题目链接
题解
orz litble
膜完题解后，突然有一个简单的想法：
考虑到(2)是质数，考虑Lucas定理：
[{n choose m} = prod_{i = 1} {lfloor frac{n}{2^{i - 1}}
floor mod 2^i choose lfloor frac{m}{2^{i - 1}}
floor mod 2^i} pmod 2
]
即
[{n choose m} = prod_{each.bit.of.n.and.m} {n' choose m'} pmod 2
]
如果二进制下有任何一位(n)为(0)且(m)不为(0)，那么就会出现(m' > n')的项，结果就为(0)了
所以结果不为(0)，当且仅当二进制下(m)是(n)的子集
所以枚举子集dp即可【(f[i])表示以(A[u] = i)的(u)开头的合法子序列个数】
([1,n])枚举子集的复杂度是(O(3^{log(max{a_i})}))
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 250000,maxm = 100005,INF = 1000000000,P = 1e9 + 7;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int f[maxn],a[maxn],ans,n;
int main(){
n = read();
REP(i,n) a[i] = read();
for (int i = n; i; i--){
int u = a[i];
for (int j = u; j; j = (j - 1) & u){
f[u] = (f[u] + f[j]) % P;
}
ans = (ans + f[u]) % P;
f[u]++;
}
printf("%d
",ans);
return 0;
}