给定大小为n的数组a，集合S包含a中所有元素，并且设x是S中的元素，那么y = 2 * x + 1, z = 4 * x都是S中的元素，求小于2^p的S的元素有多少个(mod 1e9 + 7)

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
const int maxn = 1e6 + 5, inf = 1e9 + 5, maxm = 1e6 + 5, mod = 1e9 + 7;
int a[maxn], b[maxn], c[maxn];
int n, m;
map<int, int> mp;
// bool vis[maxn];
struct Node{
    int id, val;
}suf_mx[maxn], suf_mn[maxn];

int f[maxn];
set<int> S;
bool check(int x){
    while(x){
        if(S.count(x)) return 0;
        if(x & 1){
            x = (x - 1) / 2;
        }
        else if(x % 4 == 0){
            x /= 4;
        }
        else return 1;
    }
    return 1;
}

void solve()
{
    int p;
    cin >> n >> p;
    for(int i = 0; i < p; i++){
        f[i] = 0;
    }
    S.clear();
    // a[0] = a[n + 1] = 1e9;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    sort(a + 1, a + n + 1);
    for(int i = 1; i <= n; i++){
        // cin >> a[i];
        if(check(a[i])) f[__lg(a[i])]++;
        S.insert(a[i]);
    }
    // sort(a + 1, a + n + 1);
    int res = 0;
    for(int i = 0; i < p; i++){
        if(i >= 2) f[i] = (f[i] + f[i - 1] + f[i - 2]) % mod;
        else if(i >= 1) f[i] = (f[i] + f[i - 1]) % mod;
        res = (res + f[i]) % mod;
    }
    cout << res << '\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    int T = 1;
    // cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}