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 + 10;
vector<ll > a;
ll p[N];
ll ans, MOD;

void init()
{
    int x = 0;
    while(1){
        p[x] = (1 << x) - 1;
        if(p[x] > 1e9) break;
        x++;
     }
}


void solve()
{   

    init();

    int T;
    cin >> T;
    while(T--){
       
        ans = 1;
        a.clear();

        ll n;
        cin >> n >> MOD;

        int inx = -1;
        for(int i = 1; i < 32; ++i){
            if(n >= p[i]) a.pb(p[i] - p[i - 1]);
            else{
                inx = i;
                break;
            }
        }

        if(n - p[inx - 1] > 0) a.pb(n - p[inx - 1]);
        for(auto x : a){
            ans = (ans * (x + 1)) % MOD;
        } 

        ans = (ans - 1 + MOD) % MOD;

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

int main() {

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    //cout << "ok" << endl;
    return 0;
}