POJ 3070 Fibonacci (矩阵快速幂 Fibonacci数列新求法)

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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.

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

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:

.

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MOD = 1e4;
struct matrix
{int m[5][5];
}a;matrix multiply(matrix x, matrix y)
{matrix ans;memset(ans.m, 0, sizeof(ans.m));for(int i = 1; i <= 2; i++)for(int j = 1; j <= 2; j++)if(x.m[i][j])for(int k = 1; k <= 2; k++)ans.m[i][k] = (ans.m[i][k] + x.m[i][j] * y.m[j][k]) % MOD;return ans;
}matrix quickmod(matrix a, int p)
{matrix ans;memset(ans.m, 0, sizeof(ans.m));for(int i = 1; i <= 2; i++)ans.m[i][i] = 1;while(p){if(p & 1)ans = multiply(ans, a);p >>= 1;a = multiply(a, a);}return ans;
}int main()
{   int n;matrix a;a.m[1][1] = 1; a.m[1][2] = 1;a.m[2][1] = 1;a.m[2][2] = 0;while(scanf("%d", &n) != EOF && n != -1){if(n == 0){printf("0\n");continue;}matrix ans = quickmod(a, n - 1);printf("%d\n", ans.m[1][1]);}
}