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 = 2, mod = 10000;

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

mat mul(mat A, mat B) {
    mat tp;
    for (int i = 0; i < si; i++) {
        for (int j = 0; j < si; j++) {
            tp.m[i][j] = 0;
        }
    }
    for (int i = 0; i < si; i++) {
        for (int j = 0; j < si; j++) {
            for (int k = 0; 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 = 0; i < si; i++) {
        for (int j = 0; j < si; j++) {
            if (i == j) tp.m[i][j] = 1;
            else tp.m[i][j] = 0;
        }
    }
    while (e) {
        if (e & 1) {
            tp = mul(tp, A);
        }
        A = mul(A, A);
        e /= 2;
    }
    return tp;
}
int main() {
    while (1) {
        scanf("%d", &N);
        if (N < 0) break;
        if (N == 0) {
            printf("%d\n", 0 % mod);
            continue;
        }
        mat MA;
        MA.m[0][0] = 1; MA.m[0][1] = 1;
        MA.m[1][0] = 1; MA.m[1][1] = 0;
        MA = pow(MA, N);
        printf("%d\n", MA.m[1][0] % mod);
    }
    return 0;
}