Codeforces Round #307 (Div. 2) D. GukiZ and Binary Operations

得到k二进制后，对每一位可取得的方法进行相乘即可，k的二进制形式每一位又分为2种0，1，0时，a数组必定要为一长为n的01串，且串中不出现连续的11，1时与前述情况是相反的。

且0时其方法总数为f(n) = f(n-1) + f(n-2)，其中f(2) = 3,f(1) = 3。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n,k;
int l,m;
unsigned long long p[];
queue<int> q;
//0: f[n] = f[n-2] + f[n-1]
//1: 2^i - f[n]
struct matrix{
    ll a[][];
    matrix(){
        a[][] = a[][] = a[][] = a[][] = ;
    }
    void unit(){
        a[][] = a[][] = ;
    }
    matrix operator * (const matrix& p){
        matrix ans;
        for(int i = ;i < ;++i){
            for(int j = ;j < ;++j){
                for(int k = ;k < ;++k){
                    ans.a[i][j] += a[i][k] * p.a[k][j];
                    if(ans.a[i][j] >= m) ans.a[i][j] %= m;
                }
            }
        }
        return ans;
    }
};
ll fi(){
    //f[1] = 2,f[2] = 3;
    if(n == ) return ;
    if(n == ) return ;
    ll t = n;
    t -= ;
    matrix ans,p;
    p.a[][] = ,p.a[][] = ,p.a[][] = ;
    ans.unit();
    while(t){
        if(t & ) ans = ans * p;
        p = p * p;
        t >>= ;
    }
    return ( * ans.a[][] +  * ans.a[][])%m;
}
ll quickpow(ll x,ll y){
    ll ans = ;
    while(y){
        if(y & ){
            ans = ans * x;
            if(ans >= m) ans %= m;
        }
        x *= x;
        if(x >= m) x %= m;
        y >>= ;
    }
    return ans;
}
void solve(){
    if(p[l] -  < (unsigned long long)k && l != ){
        puts("");
        return;
    }
    while(k){
        q.push(k&);
        k>>=;
    }
    ll x = fi(),y = (quickpow(,n) - x + m) % m,ans = ;
    for(int i = ;i < l;++i){
        if(!q.empty()){
            if(q.front()) ans = ans * y;
            else ans = ans * x;
            q.pop();
        }
        else{
            ans = ans * x;
        }
        if(ans >= m) ans %= m;
    }
    printf("%I64d\n",ans%m);
}
int main()
{
    cin >> n >> k >> l >> m;
    p[] = ;
    for(int i = ;i < ;++i) p[i] = p[i-]*;
    solve();
    return ;
}