Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11024 | Accepted: 7846 |
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
第一次写矩阵快速幂……留着当模板吧。
#include<iostream>
#include<cstdio>
using namespace std;
int n,a[2][2],b[2][2];
void mul(int a[2][2],int b[2][2],int ans[2][2])
{
int t[2][2];
for (int i=0;i<=1;i++)
for (int j=0;j<=1;j++)
{
t[i][j]=0;
for (int k=0;k<=1;k++)
t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000;
}
for (int i=0;i<=1;i++)
for (int j=0;j<=1;j++)
ans[i][j]=t[i][j];
}
int main()
{
while (scanf("%d",&n))
{
if (n==-1) return 0;
a[0][0]=a[0][1]=a[1][0]=1; a[1][1]=0;
b[0][0]=b[1][1]=1; b[0][1]=b[1][0]=0;
while (n)
{
if (n&1) mul(a,b,b);
n>>=1;
mul(a,a,a);
}
printf("%d\n",b[1][0]);
}
}