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
• 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

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。

 1 #include
 2 #include<string>
 3 #include
 4 #include
 5 
 6 #define ll long long
 7 #define pb push_back
 8 
 9 using namespace std;
10 
11 const int N = 1e6 + 10;
12 vector a;
13 ll p[N];
14 ll ans, MOD;
15 
16 void init()
17 {
18     int x = 0;
19     while(1){
20         p[x] = (1 << x) - 1;
21         if(p[x] > 1e9) break;
22         x++;
23      }
24 }
25 
26 
27 void solve()
28 {   
29 
30     init();
31 
32     int T;
33     cin >> T;
34     while(T--){
35        
36         ans = 1;
37         a.clear();
38 
39         ll n;
40         cin >> n >> MOD;
41 
42         int inx = -1;
43         for(int i = 1; i < 32; ++i){
44             if(n >= p[i]) a.pb(p[i] - p[i - 1]);
45             else{
46                 inx = i;
47                 break;
48             }
49         }
50 
51         if(n - p[inx - 1] > 0) a.pb(n - p[inx - 1]);
52         for(auto x : a){
53             ans = (ans * (x + 1)) % MOD;
54         } 
55 
56         ans = (ans - 1 + MOD) % MOD;
57 
58         //cout << "(ans = " << ans << ")" << endl;
59         cout << ans << endl;
60     }
61 }
62 
63 int main() {
64 
65     ios::sync_with_stdio(false);
66     cin.tie(0);
67     cout.tie(0);
68     solve();
69     //cout << "ok" << endl;
70     return 0;
71 }