H - Fibonacci POJ - 3070 （矩阵快速幂）

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, …

An alternative formula for the Fibonacci sequence is

Given an integer n, your goal is to compute the last 4 digits of Fn.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0
9
999999999
1000000000
-1
Sample Output

0
34
626
6875
Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

利用矩阵快速幂求 fibonacci 数列第n项

题解

通过设未知数计算求解常数矩阵
利用矩阵快速幂简化计算

复杂度

O(logn) == 30

数据范围

n = 1e9
每次%10000,数据不大

算法

矩阵快速幂（二分）

代码

#include<bits\stdc++.h>
using namespace std;
#define mod 10000typedef struct node {int e[3][3];
}Matrix;Matrix mp, temp, ans;
int n;void Mut(Matrix &a, Matrix &b, Matrix &c) { // 传递指针 memset(temp.e, 0, sizeof(temp.e));for(int i = 0; i <= 1; i++) {for(int j = 0; j <= 1; j++) {for(int k = 0; k <= 1; k++) {temp.e[i][j] += a.e[i][k] * b.e[k][j];temp.e[i][j] %= mod;}}}for(int i = 0; i <= 1; i++) {for(int j = 0; j <= 1; j++) {c.e[i][j] = temp.e[i][j];}}
}void quick_mod(int power) {while(power) {if(power&1) {Mut(ans, mp, ans); // 将ans*mp存到ans里 }power >>= 1;Mut(mp, mp, mp); // a=mp b=mp c=a*b }
}int main() {freopen("test.in", "r", stdin);while(cin >> n && n != -1) {if(n == 0) {printf("0\n");continue;}mp.e[0][0] = 0;mp.e[0][1] = mp.e[1][0] = mp.e[1][1] = 1;ans.e[0][0] = ans.e[0][1] = 1;quick_mod(n-1); // 求第n项，即求 （1，1）*A^(n-1) printf("%d\n", ans.e[0][0]);}return 0;
}

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H - Fibonacci POJ - 3070 （矩阵快速幂）

题解

复杂度

数据范围

算法

代码

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