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。
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct Matrix{
int m[][];
Matrix(){
memset(m, , sizeof(m));
}
}Matrix;
Matrix mtMul(Matrix A, Matrix B){
Matrix tmp;
for(int i = ;i < ;i ++)
for(int j = ;j < ;j ++)
for(int k = ;k < ;k ++){
int t = (A.m[i][k] * B.m[k][j])%;
tmp.m[i][j] = (tmp.m[i][j] + t)%;
}
return tmp;
}
Matrix mtPow(Matrix A, int k){
if(k == ) return A;
Matrix tmp = mtPow(A, k >> );
Matrix res = mtMul(tmp, tmp);
if(k & ) res = mtMul(res, A);
return res;
}
int main(){
int n;
while(~scanf("%d", &n) && (n+)){
if(n == ) printf("0\n");
else{
Matrix M;
M.m[][] = M.m[][] = M.m[][] = ;
M.m[][] = ;
Matrix tmp = mtPow(M, n);
printf("%d\n", tmp.m[][]);
}
}
return ;
}