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

2015-08-07 12:37 381 查看
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;
}


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