poj3070 Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12177 | Accepted: 8643 |
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
矩阵乘法裸题，题目已经把题解讲的很清楚了...
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define mod 10000
using namespace std;
int n;
struct matrix
{
int f[2][2];
matrix(){memset(f,0,sizeof(f));}
friend matrix operator *(const matrix &a,const matrix &b)
{
matrix c;
F(i,0,1) F(j,0,1) F(k,0,1) c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j])%mod;
return c;
}
};
int main()
{
while (scanf("%d",&n))
{
if (n==-1) break;
matrix a,b;
a.f[0][0]=a.f[0][1]=a.f[1][0]=1;a.f[1][1]=0;
b.f[1][0]=b.f[0][1]=0;b.f[0][0]=b.f[1][1]=1;
for(;n;n>>=1,a=a*a) if (n&1) b=b*a;
printf("%d\n",b.f[1][0]);
}
return 0;
}