【费马小定理降幂+矩阵快速幂+快速幂】M斐波那契数列 HDU - 4549

Think:

1知识点：费马小定理降幂+矩阵快速幂+快速幂

（１）：费马小定理降幂：

定理：若gcd(A, M) == 1，则A^x = A^(x%Eular(M))(mod M)

注：Eular(M)为M的欧拉函数值

2题意：

已知：

F[0] = a

F[1] = b

F
= F[n-1] * F[n-2] ( n > 1 )

输入a, b, n，询问F
= ?

3解题方法：

（１）：

F[0] = a

F[1] = b

F[2] = b*a

F[3] = b^(2)*a

F[4] = b^(3)*a^(2)

F[5] = b^(5)*a^(3)

.

.

.

F
= b^(fib(n-1))*a^(fib(n-2))

（２）：由于快速幂指数太高，因此需要通过费马小定理降幂

vjudge题目链接

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int mod = 1000000007;

struct Matrix{
LL v[4][4];
};

Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len);
Matrix matrix_pow(Matrix &x, int k, int Matrix_len);
LL num_pow(LL num, LL k);

int main(){
LL a, b;
int n;
while(~scanf("%lld %lld %d", &a, &b, &n)){
if(n == 0){
printf("%lld\n", a%mod);
continue;
}
if(n == 1){
printf("%lld\n", b%mod);
continue;
}
Matrix x;
x.v[0][0] = 1, x.v[0][1] = 1;
x.v[1][0] = 1, x.v[1][1] = 0;
Matrix y = matrix_pow(x, n-2, 2);

LL t1 = (y.v[0][0] + y.v[0][1])%(mod-1);
LL t2 = (y.v[1][0] + y.v[1][1])%(mod-1);

LL ans1 = num_pow(b, t1);
LL ans2 = num_pow(a, t2);
LL ans = ans1*ans2%mod;
printf("%lld\n", ans);
}
return 0;
}
Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len){
Matrix tmp;
memset(tmp.v, 0, sizeof(tmp.v));
for(int i = 0; i < Matrix_len; i++){
for(int j = 0; j < Matrix_len; j++){
for(int k = 0; k < Matrix_len; k++){
tmp.v[i][j] += (a.v[i][k]*b.v[k][j]);
tmp.v[i][j] %= (mod-1);
}
}
}
return tmp;
}
Matrix matrix_pow(Matrix &x, int k, int Matrix_len){
Matrix tmp;
memset(tmp.v, 0, sizeof(tmp.v));
for(int i = 0; i < Matrix_len; i++)
tmp.v[i][i] = 1;
while(k){
if(k & 1)
tmp = multiply(tmp, x, Matrix_len);
x = multiply(x, x, Matrix_len);
k >>= 1;
}
return tmp;
}
LL num_pow(LL num, LL k){
LL ans = 1;
while(k){
if(k & 1)
ans = ans*num%mod;
num = num*num%mod;
k >>= 1;
}
return ans;
}