3070 Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21048		Accepted: 14416

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:

.

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

 #include <stdio.h>
 #include <iostream>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <set>
 #include <sstream>
 #include <algorithm>
 int N;
 using namespace std;
 const int si = , mod = ;

 struct mat {
     int m[si][si];
 };

 mat mul(mat A, mat B) {
     mat tp;
     for (int i = ; i < si; i++) {
         for (int j = ; j < si; j++) {
             tp.m[i][j] = ;
         }
     }
     for (int i = ; i < si; i++) {
         for (int j = ; j < si; j++) {
             for (int k = ; k < si; k++) {
                 tp.m[i][j] += A.m[i][k] * B.m[k][j];
                 tp.m[i][j] %= mod;
             }
         }
     }
     return tp;
 }
 mat pow (mat A, int e){
     mat tp;
     for (int i = ; i < si; i++) {
         for (int j = ; j < si; j++) {
             if (i == j) tp.m[i][j] = ;
             else tp.m[i][j] = ;
         }
     }
     while (e) {
         if (e & ) {
             tp = mul(tp, A);
         }
         A = mul(A, A);
         e /= ;
     }
     return tp;
 }
 int main() {
     while () {
         scanf("%d", &N);
         if (N < ) break;
         if (N == ) {
             printf("%d\n",  % mod);
             continue;
         }
         mat MA;
         MA.m[][] = ; MA.m[][] = ;
         MA.m[][] = ; MA.m[][] = ;
         MA = pow(MA, N);
         printf("%d\n", MA.m[][] % mod);
     }
     return ;
 }