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.
.
题解：
按题目要求很容易写出矩阵
F[n] 0 1 1
F[n-1] 0 1 0
直接上矩阵快速幂即可
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const int mod=;
int n;
struct matrix
{
ll a[][];
matrix(){for(int i=;i<=;i++)for(int j=;j<=;j++)a[i][j]=;}
matrix(ll b[][]){for(int i=;i<=;i++)for(int j=;j<=;j++)a[i][j]=b[i][j];}
inline matrix operator *(matrix p){
matrix tmp;
for(int i=;i<=;i++)
for(int j=;j<=;j++){
tmp.a[i][j]=;
for(int k=;k<=;k++)
tmp.a[i][j]+=a[i][k]*p.a[k][j],tmp.a[i][j]%=mod;
}
return tmp;
}
};
ll work(){
if(n==)return ;
if(n==||n==)return ;
n-=;
ll t[][]={{,,},{,,},{,,}};ll sum[][]={{,,},{,,},{,,}};
matrix S=matrix(t),T=matrix(sum);
while(n){
if(n&)S=S*T;
T=T*T;n>>=;
}
return S.a[][];
}
int main()
{
while(scanf("%d",&n))
{
if(n==-)break;
printf("%lld\n",work());
}
return ;
}