【費馬小定理降冪+矩陣快速冪+快速冪】M斐波那契數列 HDU
阿新 • • 發佈:2019-01-06
Think:
1知識點:費馬小定理降冪+矩陣快速冪+快速冪
(1):費馬小定理降冪:
定理:若gcd(A, M) == 1,則A^x = A^(x%Eular(M))(mod M)
注:Eular(M)為M的尤拉函式值
2題意:
已知:
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
輸入a, b, n,詢問F[n] = ?
3解題方法:
(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[n] = b^(fib(n-1))*a^(fib(n-2))
(2):由於快速冪指數太高,因此需要通過費馬小定理降冪
以下為Accepted程式碼
#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;
}