POJ ---3070 (矩阵乘法求Fibonacci 数列)

Fibonacci

 

Description

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.

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 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).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

思路：矩阵快速幂，没什么可说的。

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct Matrix{
    int m[2][2];
    Matrix(){
        memset(m, 0, sizeof(m));
    }
}Matrix;
Matrix mtMul(Matrix A, Matrix B){
    Matrix tmp;
    for(int i = 0;i < 2;i ++)
        for(int j = 0;j < 2;j ++)
            for(int k = 0;k < 2;k ++){
                int t = (A.m[i][k] * B.m[k][j])%10000;
                tmp.m[i][j] = (tmp.m[i][j] + t)%10000;  
            }
    return tmp;
}
Matrix mtPow(Matrix A, int k){
    if(k == 1) return A;
    Matrix tmp = mtPow(A, k >> 1);
    Matrix res = mtMul(tmp, tmp);
    if(k & 1) res = mtMul(res, A);
    return res;
}
int main(){
    int n;
    while(~scanf("%d", &n) && (n+1)){
        if(n == 0) printf("0
");
        else{
            Matrix M;
            M.m[0][0] = M.m[0][1] = M.m[1][0] = 1;
            M.m[1][1] = 0;
            Matrix tmp = mtPow(M, n);
            printf("%d
", tmp.m[1][0]);
        }
    }
    return 0;
}