HDU 4565矩阵快速幂—— So Easy!

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
typedef  long long ll;
ll MOD ;

struct Matrix {
ll a[2][2];
Matrix() {
memset(a, 0, sizeof(a));
}

Matrix operator * (const Matrix y) {
Matrix ans;
for(int i = 0; i <= 1; i++)
for(int j = 0; j <= 1; j++)
for(int k = 0; k <= 1; k++)
ans.a[i][j] += a[i][k]*y.a[k][j];
for(int i = 0; i <= 1; i++)
for(int j = 0; j <= 1; j++)
ans.a[i][j] %= MOD;
return ans;
}

void operator = (const Matrix b) {
for(int i = 0; i <= 1; i++)
for(int j = 0; j <= 1; j++)
a[i][j] = b.a[i][j];
}
};

ll solve(ll a,ll b,ll n) {
Matrix ans, trs;

ans.a[0][1] = 1;
//初始值
ans.a[0][0] = a;
ans.a[0][1] = 2;

// f(n) = a*f(n-1) + b * f(n-2);
trs.a[0][0] = a;
trs.a[1][0] = -b;
trs.a[0][1] = 1;

while(n) {
if(n&1)
ans = ans*trs;
trs = trs*trs;
n >>= 1;
}
return (ans.a[0][1]+ans.a[1][1]+MOD)%MOD;//improtant
}

int main() {

ios_base::sync_with_stdio(false);
//    freopen("data.txt","r",stdin);
ll aa,bb,m,n;
while(cin>>aa>>bb>>n>>m)
{
ll p = 2*aa;
ll q = aa*aa-bb;
MOD  = m;
cout<<solve(p,q,n)<<endl;
}
return 0;
}