| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12732 | Accepted: 9060 |
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
//快速幂矩阵
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Mat{
int mat[][];
};
const int Mod = ;
Mat operator *(Mat a, Mat b)
{
Mat c;
memset(c.mat,,sizeof(c.mat));
for(int i = ; i < ; i++)
{
for(int j = ; j < ; j++)
{
for(int k = ; k < ; k++)
{
c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%Mod)%Mod;
}
}
}
return c;
}
Mat multi(int n)
{
Mat c;
memset(c.mat,,sizeof(c.mat));
c.mat[][] = c.mat[][] = ;
Mat a;
memset(a.mat,,sizeof(a.mat));
a.mat[][] = a.mat[][] = a.mat[][] = ;
while(n)
{
if(n&) c = c*a;
a = a*a;
n>>=;
}
return c;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n==-) return ;
Mat ans = multi(n);
printf("%d\n",ans.mat[][]);
}
return ;
}