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:
.
题目链接：http://poj.org/problem?id=3070
矩阵快速幂模板题
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod = 1e4;
struct matrix{
long long int m[2][2];
};
matrix multi(matrix a, matrix b){
matrix ans;
ans.m[0][0] = (a.m[0][0]*b.m[0][0]+a.m[0][1]*b.m[1][0])%mod;
ans.m[0][1] = (a.m[0][0]*b.m[0][1]+a.m[0][1]*b.m[1][1])%mod;
ans.m[1][0] = (a.m[1][0]*b.m[0][0]+a.m[1][1]*b.m[1][0])%mod;
ans.m[1][1] = (a.m[1][0]*b.m[0][1]+a.m[1][1]*b.m[1][1])%mod;
return ans;
}
int matrixpow_k(int n)
{
matrix ans, tmp;
ans.m[0][0] = ans.m[1][1] = tmp.m[0][0] = tmp.m[0][1] = tmp.m[1][0] = 1;
ans.m[0][1] = ans.m[1][0] = tmp.m[1][1] = 0;
n -= 2;
while(n)
{
if(n & 1)
ans = multi(tmp,ans);
tmp = multi(tmp, tmp);
n >>= 1;
}
int s = (ans.m[0][0]+ans.m[0][1]) % mod;
return s;
}
int main()
{
int n;
while(cin >> n)
{
if(n == -1)
break;
if(n == 0)
{
cout << 0 << endl;
continue;
}
else if(n==1 || n==2)
{
cout << 1 << endl;
continue;
}
cout << matrixpow_k(n) << endl;
}
return 0;
}