POJ 3070 Fibonacci(矩阵快速幂)
POJ 3070 Fibonacci(矩阵快速幂)
Fibonacci Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12271 Accepted: 8707 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
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
const int mod=10000;
typedef long long int LL;
LL n;
struct Node
{
int a[3][3];
};
Node multiply(Node a,Node b)
{
Node c;
memset(c.a,0,sizeof(c.a));
for(int i=0;i<2;i++)
{
for(int j=0;j<2;j++)
{
for(int k=0;k<2;k++)
{
(c.a[i][k]+=(a.a[i][j]*b.a[j][k])%mod)%=mod;
}
}
}
return c;
}
Node get(Node a,LL x)
{
Node c;
memset(c.a,0,sizeof(c.a));
for(int i=0;i<2;i++)
c.a[i][i]=1;
for(x;x;x>>=1)
{
if(x&1) c=multiply(c,a);
a=multiply(a,a);
}
return c;
}
int main()
{
while(scanf("%lld",&n)!=EOF)
{
if(n==-1)
break;
if(n==0)
{
printf("0\n");
continue;
}
Node a;
a.a[0][0]=1;a.a[0][1]=1;
a.a[1][0]=1;a.a[1][1]=0;
a=get(a,n);
printf("%d\n",a.a[0][1]);
}
return 0;
}