给定数字n。您需要找到(Fib(n)^ Fib(n))％10 ^ 9 + 7，其中Fib(n)是第n个斐波那契数。
例子 ：
Input : n = 4
Output : 27
4th fibonacci number is 3
[ fib(4) ^ fib(4) ] % 10^9 + 7 = ( 3 ^ 3 ) % 10^9 + 7 = 27
Input : n = 3
Output : 4
3th fibonacci number is 2
[ fib(3) ^ fib(3) ] % 10^9 + 7 = ( 2 ^ 2 ) % 10^9 + 7 = 4
如果n大， fib(n)将很大，并且fib(n) ^ fib(n)不仅难以计算，而且无法存储。
方法：
(a ^(p-1))％p = 1，其中p是质数(使用Fermat Little定理)。
(a ^ a)％m可以写成((a a％m)^ a)％m
也可以将任何数字“ a”写为a = k *(m – 1)+ r(使用除法算法)
其中“ k”是商，“ r”是余数。可以说r = a％(m-1)
因此，简化我们计算的步骤，假设a = fib(n)
( a ^ a ) % m
= ( ( a % m ) ^ a ) % m
= ( ( a % m ) ^ ( k * ( m - 1 ) + r ) ) % m
= ( ( ( a % m ) ^ ( m-1 ) ) ^ k * ( a % m ) ^ r ) % m
= ( (1 ^ k) * ( a % m ) ^ r ) % m
= ( ( a % m ) ^ r ) % m
= ( ( a % m ) ^ (a % (m-1) ) % m
％m和％(m-1)易于计算和存储。
并且我们可以使用此GFG文章来计算(x ^ y)％m。
使用这篇GFG文章，我们可以在log(n)时间中找到第n个斐波那契数。
下面是上述方法的实现：
C++
#include
#define mod 1000000007
using namespace std;
// Iterative function to compute modular power
long long modularexpo(long long x, long long y, long long p)
{
// Initialize result
long long res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Helper function that multiplies 2 matrices
// F and M of size 2*2, and puts the
// multiplication result back to F[][]
void multiply(long long F[2][2], long long M[2][2], long long m)
{
long long x = ((F[0][0] * M[0][0]) % m +
(F[0][1] * M[1][0]) % m) % m;
long long y = ((F[0][0] * M[0][1]) % m +
(F[0][1] * M[1][1]) % m) % m;
long long z = ((F[1][0] * M[0][0]) % m +
(F[1][1] * M[1][0]) % m) % m;
long long w = ((F[1][0] * M[0][1]) % m +
(F[1][1] * M[1][1]) % m) % m;
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
// Helper function that calculates F[][] raise to
// the power n and puts the result in F[][]
// Note that this function is designed only for fib()
// and won't work as general power function
void power(long long F[2][2], long long n, long long m)
{
if (n == 0 || n == 1)
return;
long long M[2][2] = { { 1, 1 }, { 1, 0 } };
power(F, n / 2, m);
multiply(F, F, m);
if (n % 2 != 0)
multiply(F, M, m);
}
// Function that returns nth Fibonacci number
long long fib(long long n, long long m)
{
long long F[2][2] = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1, m);
return F[0][0];
}
// Driver Code
int main()
{
long long n = 4;
long long base = fib(n, mod) % mod;
long long expo = fib(n, mod - 1) % (mod - 1);
long long result = modularexpo(base, expo, mod) % mod;
cout << result << endl;
} Java
class GFG
{
static int mod = 1000000007;
// Iterative function to compute modular power
static long modularexpo(long x, long y, long p)
{
// Initialize result
long res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Helper function that multiplies 2 matrices
// F and M of size 2*2, and puts the
// multiplication result back to F[][]
static void multiply(long F[][],
long M[][], long m)
{
long x = ((F[0][0] * M[0][0]) % m +
(F[0][1] * M[1][0]) % m) % m;
long y = ((F[0][0] * M[0][1]) % m +
(F[0][1] * M[1][1]) % m) % m;
long z = ((F[1][0] * M[0][0]) % m +
(F[1][1] * M[1][0]) % m) % m;
long w = ((F[1][0] * M[0][1]) % m +
(F[1][1] * M[1][1]) % m) % m;
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
// Helper function that calculates F[][] raise to
// the power n and puts the result in F[][]
// Note that this function is designed only for fib()
// and won't work as general power function
static void power(long F[][], long n, long m)
{
if (n == 0 || n == 1)
return;
long M[][] = { { 1, 1 }, { 1, 0 } };
power(F, n / 2, m);
multiply(F, F, m);
if (n % 2 != 0)
multiply(F, M, m);
}
// Function that returns nth Fibonacci number
static long fib(long n, long m)
{
long F[][] = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1, m);
return F[0][0];
}
// Driver Code
public static void main(String[] args)
{
long n = 4;
long base = fib(n, mod) % mod;
long expo = fib(n, mod - 1) % (mod - 1);
long result = modularexpo(base, expo, mod) % mod;
System.out.println(result);
}
}
// This code is contributed by PrinciRaj1992Python3
mod = 1000000007;
# Iterative function to compute modular power
def modularexpo(x, y, p):
# Initialize result
res = 1;
# Update x if it is more than or
# equal to p
x = x % p;
while (y > 0):
# If y is odd, multiply x with result
if (y & 1):
res = (res * x) % p;
# y must be even now
y = y >> 1;
x = (x * x) % p;
return res;
# Helper function that multiplies 2 matrices
# F and M of size 2*2, and puts the
# multiplication result back to F[][]
def multiply(F, M, m):
x = ((F[0][0] * M[0][0]) % m +
(F[0][1] * M[1][0]) % m) % m;
y = ((F[0][0] * M[0][1]) % m +
(F[0][1] * M[1][1]) % m) % m;
z = ((F[1][0] * M[0][0]) % m +
(F[1][1] * M[1][0]) % m) % m;
w = ((F[1][0] * M[0][1]) % m +
(F[1][1] * M[1][1]) % m) % m;
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
# Helper function that calculates F[][] raise to
# the power n and puts the result in F[][]
# Note that this function is designed only for fib()
# and won't work as general power function
def power(F, n, m):
if (n == 0 or n == 1):
return;
M = [[ 1, 1 ], [ 1, 0 ]];
power(F, n // 2, m);
multiply(F, F, m);
if (n % 2 != 0):
multiply(F, M, m);
# Function that returns nth Fibonacci number
def fib(n, m):
F = [[ 1, 1 ], [ 1, 0 ]];
if (n == 0):
return 0;
power(F, n - 1, m);
return F[0][0];
# Driver Code
if __name__ == '__main__':
n = 4;
base = fib(n, mod) % mod;
expo = fib(n, mod - 1) % (mod - 1);
result = modularexpo(base, expo, mod) % mod;
print(result);
# This code is contributed by 29AjayKumarC#
using System;
class GFG
{
static int mod = 1000000007;
// Iterative function to compute modular power
static long modularexpo(long x, long y, long p)
{
// Initialize result
long res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Helper function that multiplies 2 matrices
// F and M of size 2*2, and puts the
// multiplication result back to F[,]
static void multiply(long [,]F,
long [,]M, long m)
{
long x = ((F[0, 0] * M[0, 0]) % m +
(F[0, 1] * M[1, 0]) % m) % m;
long y = ((F[0, 0] * M[0, 1]) % m +
(F[0, 1] * M[1, 1]) % m) % m;
long z = ((F[1, 0] * M[0, 0]) % m +
(F[1, 1] * M[1, 0]) % m) % m;
long w = ((F[1, 0] * M[0, 1]) % m +
(F[1, 1] * M[1, 1]) % m) % m;
F[0, 0] = x;
F[0, 1] = y;
F[1, 0] = z;
F[1, 1] = w;
}
// Helper function that calculates F[,] raise to
// the power n and puts the result in F[,]
// Note that this function is designed only for fib()
// and won't work as general power function
static void power(long [,]F, long n, long m)
{
if (n == 0 || n == 1)
return;
long [,]M = { { 1, 1 }, { 1, 0 } };
power(F, n / 2, m);
multiply(F, F, m);
if (n % 2 != 0)
multiply(F, M, m);
}
// Function that returns nth Fibonacci number
static long fib(long n, long m)
{
long [,]F = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1, m);
return F[0, 0];
}
// Driver Code
public static void Main(String[] args)
{
long n = 4;
long Base = fib(n, mod) % mod;
long expo = fib(n, mod - 1) % (mod - 1);
long result = modularexpo(Base, expo, mod) % mod;
Console.WriteLine(result);
}
}
// This code is contributed by Rajput-Ji输出：
27
时间复杂度： log(n)