Codeforces Round #772 (Div. 2) D. Infinite Set

D. Infinite Set
思路
二进制下，2x+1相当于进一位然后+1，4x相当于进两位，令DP[i]表示进i位的情况数，则有递推方程dp[i] = dp[i - 1] + dp[i - 2],然后对于每个数，若k位该数能进的最大位数，则情况数为dp[0]~dp[k]的总和，用还需预处理出前缀和。然后对最开始的数组去重，接着算出每个数能进的位数，就做完了
Code

#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define debug freopen("_in.txt", "r", stdin);
// #define debug freopen("_in.txt", "r", stdin), freopen("_out.txt", "w", stdout);
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 3e5 + 10;
const ll maxm = 1e4 + 10;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;

ll n, p;
ll arr[maxn], brr[maxn], pre[maxn];
map<ll, ll> mp;
ll dp[maxn];

int main()
{
    
    
    // debug;
	
    scanf("%lld%lld", &n, &p);
	dp[0] = 1, dp[1] = 1, dp[2] = 2;
    pre[0] = 1;
    pre[1] = 2;
    for (ll i = 2; i <= p + 10; i++)
    {
    
    
        dp[i] = dp[i - 1] + dp[i - 2];
        dp[i] %= mod;
        pre[i] = pre[i - 1] + dp[i];
        pre[i] %= mod;
    }
    ll cnt = 0;
    for (ll i = 1; i <= n; i++)
    {
    
    
        scanf("%lld", &arr[i]);
        mp[arr[i]] = 1;
    }
    sort(arr + 1, arr + n + 1);
    for (ll i = 1; i <= n; i++)
    {
    
    
        ll flag = 1, num = arr[i];
        while (num)
        {
    
    

            if (num & 1)
            {
    
    
                num >>= 1;
            }
            else if (num % 4 == 0)
            {
    
    
                num >>= 2;
            }
            else
                break;
            if (mp[num])
            {
    
    
                flag = 0;
                break;
            }
        }
        if (flag)
        {
    
    
            brr[++cnt] = arr[i];
        }
    }
    
    ll ans = 0;
    for (ll i = 1; i <= cnt; i++)
    {
    
    
        ll len = 0;
        while ((1ll << len) <= brr[i])
            len++;
        if (p < len)
            continue;
        ans += pre[p - len];
        ans %= mod;
    }
    printf("%lld\n", ans);
}