[POJ](3070)Fibonacci ---矩阵快速幂与斐波那契
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16291 | Accepted: 11438 |
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<iostream> #include<cstring> #include<cstdio> const int MOD = 10000; using namespace std; typedef struct node { long long a[2][2]; }matrix; matrix matMul(matrix x,matrix y) //矩阵乘法 { matrix res; memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) { for(int j=0;j<2;j++) { for(int k=0;k<2;k++) { res.a[i][j]+=x.a[i][k]*y.a[k][j]; res.a[i][j]%=MOD; } } } return res; } long long pow(int p) //矩阵快速幂 { matrix res; matrix c; memset(c.a,0,sizeof(c.a)); memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) res.a[i][i] = 1; c.a[0][0]=1,c.a[0][1]=1; c.a[1][0]=1,c.a[1][1]=0; while(p) { if(1&p) res = matMul(res,c); c = matMul(c,c); p>>=1; } return res.a[0][1]; } int main() { int n; while(scanf("%d",&n)!=EOF) { if(n == -1) break; printf("%lld\n",pow(n)%MOD); } return 0; }