POJ 3070 Fibonacci （矩阵幂运算）

Time limit

1000 ms

Memory limit

65536 kB

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:

思路：矩阵快速幂+对10000取余。

代码:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>

using namespace std;

struct Matrix // 定义矩阵
{
int a[2][2];
Matrix()
{
memset(a,0,sizeof(a));
}
};

Matrix mul(Matrix a,Matrix b)//矩阵乘法+取余
{
Matrix ans;
for(int i=0; i<2; i++)
{
for(int k=0; k<2; k++)
{
if(a.a[i][k])
for(int j=0; j<2; j++)
{
ans.a[i][j]+=a.a[i][k]*b.a[k][j];
if(ans.a[i][j]>=10000)
ans.a[i][j]%=10000;
}
}
}
return ans;
}

int Mypow(int N)
{
Matrix mid,re;
mid.a[0][0] = mid.a[1][1] = 1;
re.a[0][0] = re.a[1][0] = re.a[0][1] = 1;
while(N)
{
if(N&1)mid = mul(mid,re);
re = mul(re,re);
N >>= 1;
}
return mid.a[0][1];
}

int main()
{
int N;
while(scanf("%d",&N) && N!=-1)
{
if(N == 0)
{
printf("0\n");
continue;
}
else if(N<3)
{
printf("1\n");
continue;
}
printf("%d\n",Mypow(N));
}
return 0;
}