Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K |
| Total Submissions: 3502 | Accepted: 2450 |
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
#include<stdio.h>
int n;
int a[4],b[4];
const int M=10000;
void cal(int n)
{
if(n==1||n==0)
{
a[3]=1;a[2]=1;a[1]=1;a[0]=0;
return ;
}
if(n&1)
{
cal(n/2);
b[3]=(a[3]*a[3]+a[2]*a[1])%M;
b[2]=(a[3]*a[2]+a[2]*a[0])%M;
b[1]=(a[1]*a[3]+a[0]*a[1])%M;
b[0]=(a[1]*a[2]+a[0]*a[0])%M;
for(int i=0;i<4;i++) a[i]=b[i];
b[3]=(a[3]+a[2])%M;
b[2]=a[3];
b[1]=(a[1]+a[0])%M;
b[0]=a[1];
for(int i=0;i<4;i++) a[i]=b[i];
}
else
{
cal(n/2);
b[3]=(a[3]*a[3]+a[2]*a[1])%M;
b[2]=(a[3]*a[2]+a[2]*a[0])%M;
b[1]=(a[1]*a[3]+a[0]*a[1])%M;
b[0]=(a[1]*a[2]+a[0]*a[0])%M;
for(int i=0;i<4;i++) a[i]=b[i];
}
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==-1) break;
if(n==0) {printf("0/n");continue;}
cal(n-1);
printf("%d/n",a[3]);
}
return 0;
}