Fibonacci_1
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
Given an integer n, your goal is to compute the last Fn mod (10^9 + 7).
Input
The input test file will contain a single line containing n (n ≤ 100).
There are multiple test cases!
Output
For each test case, print the Fn mod (10^9 + 7).
解题思路:一般斐波那契数列的求解有多种,下面通过递归和幂次数实现
#include <math.h> using namespace std; int Fibonacci_1(int n) { if(n==0 || n== 1)return n; int fn = Fibonacci_1(n-1) +Fibonacci_1(n-2); return fn %((int)pow(10,9)+7); //每次都进行取模,防止数值过大溢出 } int main(int argc, const char * argv[]) { int n; while(cin >> n) { cout << Fibonacci_1(n) << endl; } return 0; }
Fibonacci_2
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
Given an integer n, your goal is to compute the last Fn mod (10^9 + 7).
Input
The input test file will contain a single line containing n (n ≤ 2^31-1).
There are multiple test cases!
Output
For each test case, print the Fn mod (10^9 + 7).
解题思路:相对于Fibonacci_1,不同的是,输入的n的范围值大了很多(n ≤ 2^31-1).考虑到时间限制,需要使用幂次方矩阵来解决这个问题。
斐波那契数列用矩阵表示如下:
#include <iostream> using namespace std; #define M 1000000007 struct Matrix{ long long v[2][2]; }; Matrix matrixMul(Matrix a, Matrix b) { //两个矩阵相乘 Matrix temp; for (int i = 0; i != 2; i++) { for (int j = 0; j != 2; j++) { temp.v[i][j] = 0; for (int k = 0; k != 2; k++) { temp.v[i][j] += a.v[i][k] * b.v[k][j]; temp.v[i][j] %= M; } } } return temp; } Matrix power(Matrix a, Matrix b, long long n) { while (n) { if (n & 1) { b = matrixMul(b, a); } n >>= 1; a = matrixMul(a, a); } return b; } int main(int argc, char* argv[]) { Matrix a = {1, 1, 1, 0}, b = {1, 0, 0, 1}; long long n; while (cin >> n) { if (n == 0) cout << 0 << endl; else { Matrix result = power(a, b, n - 1); cout << result.v[0][0] << endl; } } return 0; }
1.刚开始的时候,并没有注意到数列递增的趋势,使用int是无法表示全部所需数列值的。因此后面统一变成了long long型
2.幂次方解决方法能够大大提高速度。
代码新手,有什么建议和意见欢迎提出