http://poj.org/problem?id=3070
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:
这道题就是快速幂http://blog.csdn.net/u013795055/article/details/38599321
还有矩阵相乘
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <queue>using namespace std;
#define memset(a,b) memset(a,b,sizeof(a))
#define N 4
#define INF 0xfffffff
struct node
{
int a[N][N];
}e;node mm(node p,node q)
{
node t;
for(int i=;i<;i++)
{
for(int j=;j<;j++)
{
t.a[i][j]=;
for(int k=;k<;k++)
{
t.a[i][j]=(t.a[i][j]+(p.a[i][k]*q.a[k][j]))%;
}
}
}
return t;
}node mul(node p,int n)
{ node q;
for(int i=;i<;i++)
{
for(int j=;j<;j++)
{
q.a[i][j]=(i==j);
}
}///这个是为了当n是奇数是第一次跟q相乘是还是原来的p,为了下一次跟下一奇数相乘
while(n)
{
if(n&)
q=mm(q,p);
n=n/;
p=mm(p,p);
}
return q;
}
int main()
{
int n;
while(scanf("%d",&n),n!=-)
{
e.a[][]=;
e.a[][]=;
e.a[][]=;
e.a[][]=;
node b=mul(e,n);
printf("%d\n",b.a[][]);
}
return ;
}