【POJ 3070】Fibonacci(矩阵快速幂)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12333 | Accepted: 8752 |
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
矩快基础题,训练计划里的居然一直没看。。。或者是对数论优先忽略了……
贴上来防止时间久了忘掉,还能过来看看
代码如下:
#include <iostream> #include <cmath> #include <vector> #include <cstdlib> #include <cstdio> #include <cstring> #include <queue> #include <stack> #include <list> #include <algorithm> #include <map> #include <set> #define LL long long #define Pr pair<int,int> #define fread() freopen("in.in","r",stdin) #define fwrite() freopen("out.out","w",stdout) using namespace std; const int INF = 0x3f3f3f3f; const int msz = 10000; const int mod = 1e4; const double eps = 1e-8; int ans[2][2]; int a[2][2]; int tmp[2][2]; void pow_m(int b) { for(int i = 0; i < 2; ++i) for(int j = 0; j < 2; ++j) { ans[i][j] = !(i^j); a[i][j] = !(i&j); } while(b) { if(b&1) { memset(tmp,0,sizeof(tmp)); for(int i = 0; i < 2; ++i) for(int j = 0; j < 2; ++j) for(int k = 0; k < 2; ++k) tmp[i][j] = (tmp[i][j] + ans[i][k]*a[k][j])%mod; memcpy(ans,tmp,sizeof(tmp)); } b >>= 1; memset(tmp,0,sizeof(tmp)); for(int i = 0; i < 2; ++i) for(int j = 0; j < 2; ++j) for(int k = 0; k < 2; ++k) tmp[i][j] = (tmp[i][j] + a[i][k]*a[k][j])%mod; memcpy(a,tmp,sizeof(tmp)); } } int main() { //fread(); //fwrite(); int n; while(~scanf("%d",&n) && n != -1) { pow_m(n); printf("%d\n",ans[0][1]); } return 0; }