poj 3070 Fibonacci

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

099999999991000000000-1

Sample Output

0346266875

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:.

【题意】给你一个数n，让你求斐波那契数列的第n项。

【分析】题目给的很明显了，自然是用矩阵快速幂，直接套模板就行了。

【代码】

#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <list>#include <algorithm>#include <iostream>using namespace std;const int len=2;const int mod = 10000;struct Mat {    int a[len][len];};Mat operator * (Mat a,Mat b) {    Mat c;    memset(c.a,0,sizeof(c.a));    for(int i=0; i<len; i++)        for(int k=0; k<len; k++)            for(int j=0; j<len; j++)                    c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mod;    return c;}Mat operator ^ (Mat a,int k) {    Mat c;    for(int i=0; i<len; i++)        for(int j=0; j<len; j++)            c.a[i][j]=(i==j);    while(k) {        if(k&1) {            c=c*a;        }        k>>=1;        a=a*a;    }    return c;}int main() {    int k;    while(scanf("%d",&k)!=EOF&&k!=-1) {        Mat ans;        ans.a[0][0]=ans.a[0][1]=ans.a[1][0]=1;        ans.a[1][1]=0;        ans=ans^k;        printf("%d\n",ans.a[0][1]%mod);    }    return 0;}