POJ 3070 Fibonacci【矩阵连乘】
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6881 | Accepted: 4873 |
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
#include<cstdio> const int mod = 10000; struct Matrix { __int64 a11, a12, a21, a22; }matrix; Matrix mull(Matrix a, Matrix b) { Matrix c; c.a11 = a.a11*b.a11+a.a12*b.a21; c.a12 = a.a11*b.a12+a.a12*b.a22; c.a21 = a.a21*b.a11+a.a22*b.a21; c.a22 = a.a21*b.a12+a.a22*b.a22; c.a11 %= mod; c.a12 %= mod; c.a21 %= mod; c.a22 %= mod; return c; } Matrix find(Matrix m, __int64 n) { Matrix b; b.a11 = 1; b.a12 = 0; b.a21 = 0; b.a22 = 1; while(n > 0) { if(n&1) { b = mull(b, m); } n = n>>1; m = mull(m, m); } return b; } int main() { __int64 n; while(scanf("%I64d", &n) != EOF) { if(n == -1) break; else if(n == 0) { printf("0\n"); continue; } __int64 a11, a12, a21, a22; Matrix m, ans; m.a11 = 1; m.a12 = 1; m.a21 = 1; m.a22 = 0; ans = find(m, n); __int64 result = ans.a12%10000; printf("%I64d\n", result); } return 0; }