51nod 1436 方程的解数

https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1436

大概就是一个按位分析k再加上快速幂就可以了

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const ll mod = 1000000007 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  2e5+20;
const int maxm = 3e2+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
int in(int &ret) {
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}

ll mo;
struct mat
{
    ll m[2][2];
    void init()
    {
        mst(m,0);
        m[0][0]= m[0][1] = m[1][0] = 1;
    }
    void clr()
    {
        mst(m,0);
        r0(i,2)m[i][i] = 1;
    }
    void con()
    {
        mst(m,0);
        m[0][0] = 1;
        m[1][0] = 1;
    }
    mat operator *(const mat o)const
    {
        mat t;
        mst(t.m,0);
        r0(i,2)r0(j,2)r0(k,2)
        {
            t.m[i][j] += m[i][k]*o.m[k][j]%mo;
            t.m[i][j] %= mo;
        }
        return t;
    }
};
ll qpow(ll x,ll k)
{
    ll res = 1;
    while(k)
    {
        if(k&1)res = res*x%mo;
        x = x*x%mo;
        k>>=1;
    }
    return res;
}
ll cntbit(ll x)
{
    int cb = -1;
    while(x)
    {
        x>>=1;
        ++cb;
    }
    return cb;
}
ll calfib(ll k)
{
    mat base,first,res;
    res.con();
    base.init();
    first.clr();
    while(k)
    {
        if(k&1)first = first * base;
        base = base *base;
        k>>=1;
    }
    res = first *res;
    return res.m[0][0]%mo;
}
int main() {
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    ll n,k,l;
    slddd(n,k,l);
    sld(mo);
    if(mo==1||cntbit(k)>=l)
    {
        return puts("0")*0;
    }
    ll x = calfib(n);
    ll y = qpow(2,n) - x;
    y = (y+mo)%mo;
    int p = 0,q = 0;
    while(k)
    {
        if(k&1)++p;
        else ++q;
        k>>=1;
    }
    q += l-p-q;
    ll ans = 1;
    ans = ans * qpow(x,q)%mo;
    ans = ans * qpow(y,p)%mo;
    lansn();
    return 0;
}