| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9517 | Accepted: 6767 |
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
题目中给了提示矩阵。
网赛时才接触矩阵幂运算，自己掌握的知识还是太少了。。。。
代码：
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=2;
const int mod=10000;
int n=2;//矩阵大小
int num;//幂大小
struct mat
{
int arr[maxn][maxn];
mat()
{
memset(arr,0,sizeof(arr));
}
};
mat mul(mat a,mat b)
{
mat ans;
for(int i=0;i<n;i++)
{
for(int k=0;k<n;k++)
{
if(a.arr[i][k])
for(int j=0;j<n;j++)
{
ans.arr[i][j]+=a.arr[i][k]*b.arr[k][j];
if(ans.arr[i][j]>=mod)
ans.arr[i][j]%=mod;
}
}
}
return ans;
}
mat power(mat p,int k)
{
if(k==1) return p;
mat e;
for(int i=0;i<n;i++)
e.arr[i][i]=1;
if(k==0) return e;
while(k)
{
if(k&1)
e=mul(e,p);
p=mul(p,p);
k>>=1;
}
return e;
}
int main()
{
mat ori;//题目中的原始矩阵
ori.arr[0][0]=ori.arr[0][1]=ori.arr[1][0]=1;
while(cin>>num&&num!=-1)
{
mat ans;
ans=power(ori,num);
cout<<ans.arr[0][1]<<endl;
}
return 0;
}
[ACM] POJ 3070 Fibonacci （矩阵幂运算）
原文地址：http://blog.csdn.net/sr_19930829/article/details/39368025
