POJ 3070 矩阵快速幂

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:



想法：矩阵做一次乘法需要三重循环，而题目中的n给得又是那么得大，所以用暴力的解法肯定是TLE的，所以我们想到了快速幂的方法，只要把其中的乘法换成矩阵的乘法法则即可。

//base的n次方.快速幂方式
int fast_mod(int n,int base)
{
int ans = 1;
while(n)
{
if(n&1)  ans = ans*base%mod;
base = base*base%mod;
n>>=1;
}
return ans;
}

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <map>
#include <iostream>
using namespace std;
long long phi[3000010];
const int mod = 10000;
struct matrix
{
int m[2][2];
}a,base;
matrix xiangcheng(matrix a,matrix base)
//矩阵乘法的代码
{
matrix temp;
for (int i = 0; i<2; i++) {
for (int j = 0; j<2; j++) {
temp.m[i][j] = 0;
for (int k = 0; k<2; k++) {
temp.m[i][j] =(temp.m[i][j]+a.m[i][k]*base.m[k][j])%mod;
}
}
}
return temp;
}
int fast_mod(long long n)
{
base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
base.m[1][1] = 0;
//初始化base.
a.m[0][0] = a.m[1][1] = 1;
a.m[0][1] = a.m[1][0] = 0;
//把a初始化成单位矩阵。
while (n) {
//快速幂
if (n&1) {
//判断最后一位是否为1.
a = xiangcheng(a, base);
//换成矩阵乘法.
}
base = xiangcheng(base, base);
n>>=1;
//左移一位.
}
return a.m[0][1];
}
int main()
{
long long n;
cin>>n;
while (n!=-1) {
cout<<fast_mod(n)%mod<<endl;
cin>>n;
}
return 0;
}