hdu 5667 Sequence(矩阵快速幂)

Problem Description

Holion
August will eat every thing he has found.

Now
there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise

He
gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But
there are only p foods,so you should tell him fn mod
p.

Input

The
first line has a number,T,means testcase.

Each
testcase has 5 numbers,including n,a,b,c,p in a line.

1≤T≤10,1≤n≤1018,1≤a,b,c≤109,p is
a prime number,and p≤109+7.

Output

Output
one number for each case,which is fn mod
p.

Sample Input

1
5 3 3 3 233

Sample Output

190

solution：
对于递推式，取以a为底的对数，得到f
=c*f[n-1]+f[n-2]+b;此时用矩阵快速幂即可。
tips:
因为p为素数，且模的位置在指数，因此模p-1即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 5;
typedef long long ll;
struct Mat{
int row, col;
ll mat[maxn][maxn];
};
Mat mat_mul(Mat  A, Mat B, ll p)
{
Mat ans;
ans.row = A.row;
ans.col = B.col;
memset(ans.mat, 0, sizeof(ans.mat));
for (int i = 0; i < A.row; i++)
for (int j = 0; j < A.col; j++)
if (A.mat[i][j])
for (int k = 0; k < B.col; k++)
{
ans.mat[i][k] += A.mat[i][j] * B.mat[j][k];
ans.mat[i][k] %= p;
}
return ans;
}
Mat mat_pow(Mat A,ll x,ll p)
{
Mat C;
C.col = A.row;
C.row = A.col;
for (int i = 0; i < C.row; i++)
for (int j = 0; j < C.col; j++)
C.mat[i][j] = (i == j);
while (x)
{
if (x & 1)C = mat_mul(C, A, p);
A = mat_mul(A, A, p);
x >>= 1;
}
return C;
}
ll pow(ll x, ll y, ll p)
{
ll ans = 1;
while (y)
{
if (y & 1)ans = (ans*x) % p;
x = (x*x) % p;
y >>= 1;
}
return ans%p;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
ll n, a, b, c, p;
scanf("%I64d%I64d%I64d%I64d%I64d", &n, &a, &b, &c, &p);
Mat A, B;
A.row = A.col = 3;
A.mat[0][0] = c; A.mat[0][1] = 1; A.mat[0][2] = 1;
A.mat[1][0] = 1; A.mat[1][1] = 0; A.mat[1][2] = 0;
A.mat[2][0] =0; A.mat[2][1] = 0; A.mat[2][2] = 1;
if (n == 1)printf("1\n");
else if (n == 2)printf("%I64d\n", pow(a, b, p));
else {
B = mat_pow(A,n - 2,p-1);
ll ans = B.mat[0][0] * b + B.mat[0][2]*b;
printf("%I64d\n", pow(a, ans, p));
}
}
}