poj 3070-Fibonacci-矩阵幂乘

Fibonacci

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3070
64-bit integer IO format: %lld      Java class name: Main

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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

099999999991000000000-1

Sample Output

0346266875

代码1：

#include<stdio.h>#include<string.h>#include<math.h>int m[3][3];int a[100050];void f() {a[0] = 0;a[1] = 1;for (int i  = 2; i <= 100050; i ++) {a[i] = (a[i - 1] + a[i - 2]) % 10000;}}int main () {int n;while (scanf("%d", &n) != EOF) {if(n == -1) break;f();printf("%d\n", a[n % 15000]);}return 0;}