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Fibonacci
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Time Limit: 1000MS |
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Memory Limit: 65536K |
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Total Submissions: 13172 |
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Accepted: 9368 |
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
思路：没得说，矩阵快速幂
代码如下：
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod = ;
const int N = ;//矩阵的维数,角标从0开始
struct Matrix
{
__int64 v[N][N];
Matrix()
{
memset(v,,sizeof(v));
}
};
//矩阵的乘法p1*p2
Matrix multi(Matrix p1,Matrix p2)
{
Matrix res;
for(int i=;i<N;i++)
for(int j=;j<N;j++)
if(p1.v[i][j])//代码优化，是0的话就不用计算
for(int k=;k<N;k++)
res.v[i][k]=(res.v[i][k]+(p1.v[i][j]*p2.v[j][k]))%mod;
return res;
}
//矩阵的快速幂p^k
Matrix pow(Matrix p,__int64 k)
{
Matrix t;
for(int i=;i<N;i++)//初始化为单位矩阵
t.v[i][i]=;
while(k)
{
if(k&)
t=multi(t,p);
p=multi(p,p);
k=k>>;
}
return t;
}
int main()
{
__int64 n;
Matrix e,ans;
e.v[][]=e.v[][]=e.v[][]=;
e.v[][]=;
while(scanf("%I64dd",&n)!=EOF&&n!=-)
{
ans = pow(e,n);
printf("%I64d\n",ans.v[][]);
}
return ;
}
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