hpuvj【3070】Fibonacci（矩阵快速幂求模）

Fibonacci

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

一道涉及到矩阵快速幂的知识的题 本来快速幂没学好 今天还专门说这个了，所以 看了半天才略懂了 一些，代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Mat//定义矩阵
{
int a[2][2];
void init()
{
memset(a,0,sizeof(a));
a[0][0]=a[1][1]=1;
}
};
Mat mul(Mat a,Mat b)//矩阵乘法
{
Mat ans;
ans.init();
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
ans.a[i][j]=0;
for(int k=0;k<2;k++)
{
ans.a[i][j]+=a.a[i][k]*b.a[k][j];
}
ans.a[i][j]%=10000;
}
}
return ans;
}
Mat power(Mat a,int num)//矩阵快速幂模板
{
Mat ans;
ans.init();
while(num)
{
if(num%2==1)
{
ans=mul(ans,a);
}
num/=2;
a=mul(a,a);
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF&&n!=-1)
{
if(n==0)
{
printf("0\n");
continue;
}
Mat a,ans;
a.a[0][1]=a.a[0][0]=a.a[1][0]=1;
a.a[1][1]=0;
ans=power(a,n);
printf("%d\n",ans.a[0][1]%10000);
}
return 0;
}

Fibonacci

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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