poj3070 矩阵快速幂

Posted 2020-08-11

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13438		Accepted: 9562

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

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:

.

第一次写快速矩阵幂，按着Hint讲解的就好， 联想着快速幂，都有一个幂n,底数a,取余mod，还有一个ans = 1，作为初始；这里用

代替ans = 1， [1,1 ]作为底数a。。大概如此。。

                    [ 1,0]

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 const int mod = 10000;
 5 using namespace std;
 6 struct mat {
 7     int a[3][3];
 8     mat() {
 9         memset(a, 0, sizeof(a));
10     }
11     mat operator *(const mat &o) const {
12         mat t;
13         for(int i = 1; i <= 2; i++) {
14             for(int j = 1; j <= 2; j++)
15                 for(int k = 1; k <= 2; k++) {
16                     t.a[i][j] = (t.a[i][j] + a[i][k]*o.a[k][j])%mod;
17                 }
18         }
19         return t;
20     }
21 } a, b;
22 
23 int main() {
24     int n;
25     while(scanf("%d", &n) !=EOF && n!= -1) {
26         a.a[1][1] = a.a[2][2] = 1, a.a[1][2] = a.a[2][1] = 0;
27         b.a[1][1] = b.a[1][2] = b.a[2][1] = 1, b.a[2][2] = 0;
28         if(n == 0) printf("0\n");
29         else {
30             n--;
31             while(n > 0) {
32                 if(n&1) {
33                     a = b*a;
34                     n--;
35                 }
36                 n >>= 1;
37                 b = b*b;
38             }
39             printf("%d\n", a.a[1][1]);
40         }
41     }
42     return 0;
43 }