Codeforces 551D GukiZ and Binary Operations(矩阵快速幂)

Problem D. GukiZ and Binary Operations

Solution 

      一位一位考虑，就是求一个二进制序列有连续的1的种类数和没有连续的1的种类数。

     没有连续的1的二进制序列的数目满足f[i]=f[i-1]+f[i-2],恰好是斐波那契数列。

     数据范围在10^18，用矩阵加速计算，有连续的1的数目就用2^n-f[n+1]

     最后枚举k的每一位,是1乘上2^n-f[n+1],是0乘上f[n+1]

     注意以上需要满足 2^l>k。并且这里l的最大值为64，需要特判。

 

#include <bits/stdc++.h>



using namespace std;



typedef unsigned long long ll;

const int N = 2;

ll n, k, l, m;



struct Mat {

    ll mat[N + 1][N + 1];

} A, B;



Mat operator * ( Mat a, Mat b )

{

    Mat c;

    memset ( c.mat, 0, sizeof c.mat );

    for ( int k = 0; k <  N; k++ )

        for ( int i = 0; i <  N; i++ )

            for ( int j = 0; j <  N; j++ )

                ( c.mat[i][j] += ( a.mat[i][k] * b.mat[k][j] ) % m ) %= m;

    return c;

}



Mat operator ^ ( Mat a, ll pow )

{

    Mat c;

    for ( int i = 0; i <  N; i++ )

        for ( int j = 0; j <  N; j++ )

            c.mat[i][j] = ( i == j );

    while ( pow ) {

        if ( pow & 1 )     c = c * a;

        a = a * a;

        pow >>= 1;

    }

    return c;

}



ll quickp( ll x )

{

    ll s = 1, c = 2;

    while( x ) {

        if( x & 1 ) s = ( s * c ) % m;

        c = ( c * c ) % m;

        x >>= 1;

    }

    return s;

}

int main()

{

    ios::sync_with_stdio( 0 );

    Mat p, a;

    p.mat[0][0] = 0, p.mat[0][1] = 1;

    p.mat[1][0] = 1, p.mat[1][1] = 1;

    a.mat[0][0] = 1, a.mat[0][1] = 2;

    a.mat[1][0] = 0, a.mat[1][1] = 0;

    cin >> n >> k >> l >> m;



    ll ans = 0;

    if(  l == 64 || ( 1uLL << l ) > k  ) {

        ans++;

        p = p ^ n;

        a = a * p;

        ll t1 = a.mat[0][0], t2 = ( m + quickp( n ) - t1 ) % m;

        for( int i = 0; i < l; ++i ) {

            if( k & ( 1uLL << i ) ) ans = ( ans * t2 ) % m;

            else ans = ( ans * t1 ) % m;

        }

    }



    cout << ans%m << endl;



}

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