【POJ 3070】Fibonacci（矩阵快速幂）

【POJ 3070】Fibonacci（矩阵快速幂）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12333		Accepted: 8752

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:


.

Source
Stanford Local 2006

矩快基础题，训练计划里的居然一直没看。。。或者是对数论优先忽略了……

贴上来防止时间久了忘掉，还能过来看看

代码如下:

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e4;
const double eps = 1e-8;

int ans[2][2];
int a[2][2];
int tmp[2][2];

void pow_m(int b)
{
for(int i = 0; i < 2; ++i)
for(int j = 0; j < 2; ++j)
{
ans[i][j] = !(i^j);
a[i][j] = !(i&j);
}

while(b)
{
if(b&1)
{
memset(tmp,0,sizeof(tmp));
for(int i = 0; i < 2; ++i)
for(int j = 0; j < 2; ++j)
for(int k = 0; k < 2; ++k)
tmp[i][j] = (tmp[i][j] + ans[i][k]*a[k][j])%mod;
memcpy(ans,tmp,sizeof(tmp));
}

b >>= 1;

memset(tmp,0,sizeof(tmp));
for(int i = 0; i < 2; ++i)
for(int j = 0; j < 2; ++j)
for(int k = 0; k < 2; ++k)
tmp[i][j] = (tmp[i][j] + a[i][k]*a[k][j])%mod;
memcpy(a,tmp,sizeof(tmp));
}
}

int main()
{
//fread();
//fwrite();

int n;
while(~scanf("%d",&n) && n != -1)
{
pow_m(n);
printf("%d\n",ans[0][1]);
}

return 0;
}