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
思路：矩阵快速幂，没什么可说的。
1 #include<cstdio> 2 #include<string> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 typedef struct Matrix{ 8 int m[2][2]; 9 Matrix(){ 10 memset(m, 0, sizeof(m)); 11 } 12 }Matrix; 13 Matrix mtMul(Matrix A, Matrix B){ 14 Matrix tmp; 15 for(int i = 0;i < 2;i ++) 16 for(int j = 0;j < 2;j ++) 17 for(int k = 0;k < 2;k ++){ 18 int t = (A.m[i][k] * B.m[k][j])%10000; 19 tmp.m[i][j] = (tmp.m[i][j] + t)%10000; 20 } 21 return tmp; 22 } 23 Matrix mtPow(Matrix A, int k){ 24 if(k == 1) return A; 25 Matrix tmp = mtPow(A, k >> 1); 26 Matrix res = mtMul(tmp, tmp); 27 if(k & 1) res = mtMul(res, A); 28 return res; 29 } 30 int main(){ 31 int n; 32 while(~scanf("%d", &n) && (n+1)){ 33 if(n == 0) printf("0 "); 34 else{ 35 Matrix M; 36 M.m[0][0] = M.m[0][1] = M.m[1][0] = 1; 37 M.m[1][1] = 0; 38 Matrix tmp = mtPow(M, n); 39 printf("%d ", tmp.m[1][0]); 40 } 41 } 42 return 0; 43 }