CF 1329B Dreamoon Likes Sequences

传送门

题目：

Dreamoon likes sequences very much. So he created a problem about the sequence that you can't find in OEIS:

You are given two integers d,m find the number of arrays a, satisfying the following constraints:

• The length of a is n, n≥1
• 1≤a1<a2<⋯<an≤d
• Define an array b of length n as follows: b1=a1, ∀i>1,bi=bi−1⊕ai, where ⊕ is the bitwise exclusive-or (xor). After constructing an array b, the constraint b1<b2<⋯<bn−1<bn should hold.

Since the number of possible arrays may be too large, you need to find the answer modulo m.

思路：容易想到如果两个数二进制的最高位都是1，则两者"^"后者数值一定比前者小，可以推断出"2^0 ~ 2^1 - 1","2^1 ~ 2^2 - 1",...,"2^(n-1) ~ 2^(n) - 1"为不同集合，我们只需要统计出每个集合的个数，然后求出组合方案数就行。

 #include<iostream>
 #include<string>
 #include<vector>
 #include<cstdio>

 #define ll long long
 #define pb push_back

 using namespace std;

 const int N = 1e6 + ;
 vector<ll > a;
 ll p[N];
 ll ans, MOD;

 void init()
 {
     int x = ;
     while(){
         p[x] = ( << x) - ;
         if(p[x] > 1e9) break;
         x++;
      }
 }

 void solve()
 {   

     init();

     int T;
     cin >> T;
     while(T--){

         ans = ;
         a.clear();

         ll n;
         cin >> n >> MOD;

         int inx = -;
         for(int i = ; i < ; ++i){
             if(n >= p[i]) a.pb(p[i] - p[i - ]);
             else{
                 inx = i;
                 break;
             }
         }

         if(n - p[inx - ] > ) a.pb(n - p[inx - ]);
         for(auto x : a){
             ans = (ans * (x + )) % MOD;
         } 

         ans = (ans -  + MOD) % MOD;

         //cout << "(ans = " << ans << ")" << endl;
         cout << ans << endl;
     }
 }

 int main() {

     ios::sync_with_stdio(false);
     cin.tie();
     cout.tie();
     solve();
     //cout << "ok" << endl;
     return ;
 }