|
Fibonacci
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:
Source |
[Submit] [Go Back] [Status] [Discuss]
通过这道题，练习了下矩阵快速幂的做法，还有学到了斐波那契数列的快速求法
AC代码:
.
.
.#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MOD = 10000;
struct matrix { //矩阵
int m[2][2];
}ans;
matrix base = {1, 1, 1, 0};
matrix multi(matrix a, matrix b) { //矩阵相乘，返回一个矩阵
matrix tmp;
for(int i = 0; i < 2; i++) {
for(int j = 0; j < 2; j++) {
tmp.m[i][j] = 0;
for(int k = 0; k < 2; k++)
tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
}
}
return tmp;
}
int matrix_pow(matrix a, int n) { //矩阵快速幂，矩阵a的n次幂
ans.m[0][0] = ans.m[1][1] = 1; //初始化为单位矩阵
ans.m[0][1] = ans.m[1][0] = 0;
while(n) {
if(n & 1) ans = multi(ans, a);
a = multi(a, a);
n >>= 1;
}
return ans.m[0][1];
}
int main() {
int n;
while(scanf("%d", &n), n != -1) {
printf("%d\n", matrix_pow(base, n));
}
return 0;
}
POJ - 3070 - Fibonacci （矩阵快速幂 + 斐波那契数列）
原文：http://blog.csdn.net/u014355480/article/details/44659245