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[][];
     Matrix(){
         memset(m, , sizeof(m));
     }
 }Matrix;
 Matrix mtMul(Matrix A, Matrix B){
     Matrix tmp;
     for(int i = ;i < ;i ++)
         for(int j = ;j < ;j ++)
             for(int k = ;k < ;k ++){
                 int t = (A.m[i][k] * B.m[k][j])%;
                 tmp.m[i][j] = (tmp.m[i][j] + t)%;
             }
     return tmp;
 }
 Matrix mtPow(Matrix A, int k){
     if(k == ) return A;
     Matrix tmp = mtPow(A, k >> );
     Matrix res = mtMul(tmp, tmp);
     if(k & ) res = mtMul(res, A);
     return res;
 }
 int main(){
     int n;
     while(~scanf("%d", &n) && (n+)){
         if(n == ) printf("0\n");
         else{
             Matrix M;
             M.m[][] = M.m[][] = M.m[][] = ;
             M.m[][] = ;
             Matrix tmp = mtPow(M, n);
             printf("%d\n", tmp.m[][]);
         }
     }
     return ;
 }