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

转自http://www.cnblogs.com/keam37/p/4578568.html

/*Author :usedrose  */
/*Created Time :2015/8/7 12:26:39*/
/*File Name :2.cpp*/
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <vector>
#include <string>
#include <ctime>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
#define MAXN 1110
#define MAXM 100110
#define OK cout << "ok" << endl;
#define o(a) cout << #a << " = " << a << endl
#define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
using namespace std;
typedef 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()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    cin.tie(0);
    ios::sync_with_stdio(false);
    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 ((1uLL<<i)&k) 
                ans = (ans*t2)%m;
            else ans = (ans*t1)%m;
        }
    }
    cout << ans%m << endl;

    return 0;
}