3070 Fibonacci
作者：互联网
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21048 | Accepted: 14416 |
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:
.
裸的快速矩阵幂 >>=1就是/=2 前几天做题目头脑没转过弯来 好气
讲解参考 https://www.cnblogs.com/cmmdc/p/6936196.html
1 #include <stdio.h>
2 #include <iostream>
3 #include <cstring>
4 #include <vector>
5 #include <queue>
6 #include <set>
7 #include <sstream>
8 #include <algorithm>
9 int N;
10 using namespace std;
11 const int si = 2, mod = 10000;
12
13 struct mat {
14 int m[si][si];
15 };
16
17 mat mul(mat A, mat B) {
18 mat tp;
19 for (int i = 0; i < si; i++) {
20 for (int j = 0; j < si; j++) {
21 tp.m[i][j] = 0;
22 }
23 }
24 for (int i = 0; i < si; i++) {
25 for (int j = 0; j < si; j++) {
26 for (int k = 0; k < si; k++) {
27 tp.m[i][j] += A.m[i][k] * B.m[k][j];
28 tp.m[i][j] %= mod;
29 }
30 }
31 }
32 return tp;
33 }
34 mat pow (mat A, int e){
35 mat tp;
36 for (int i = 0; i < si; i++) {
37 for (int j = 0; j < si; j++) {
38 if (i == j) tp.m[i][j] = 1;
39 else tp.m[i][j] = 0;
40 }
41 }
42 while (e) {
43 if (e & 1) {
44 tp = mul(tp, A);
45 }
46 A = mul(A, A);
47 e /= 2;
48 }
49 return tp;
50 }
51 int main() {
52 while (1) {
53 scanf("%d", &N);
54 if (N < 0) break;
55 if (N == 0) {
56 printf("%d\n", 0 % mod);
57 continue;
58 }
59 mat MA;
60 MA.m[0][0] = 1; MA.m[0][1] = 1;
61 MA.m[1][0] = 1; MA.m[1][1] = 0;
62 MA = pow(MA, N);
63 printf("%d\n", MA.m[1][0] % mod);
64 }
65 return 0;
66 }
标签：MA,mat,int,tp,3070,si,Fibonacci,include 来源： https://www.cnblogs.com/smatrchen/p/10592735.html