poj3070Fibonacci(矩阵快速幂)

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_{n − 1} + F_{n − 2} forn ≥ 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 F_n.

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 F_n. If the last four digits ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_n 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:

    .                 

#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;int m=10000;struct node{    int m[2][2];}res,x;typedef struct node NODE;NODE mult(NODE a,NODE b){    NODE tmp;    for(int i=0;i<2;i++)        for(int j=0;j<2;j++)    {        tmp.m[i][j]=0;        for(int k=0;k<2;k++)            tmp.m[i][j]=(tmp.m[i][j]+a.m[i][k]*b.m[k][j])%m;    }    return tmp;}int pow_mod(int n){    res.m[0][0]=res.m[1][1]=1;    res.m[0][1]=res.m[1][0]=0;    x.m[0][1]=x.m[0][0]=x.m[1][0]=1;    x.m[1][1]=0;    while(n)    {        if(n&1)        {            res=mult(res,x);        }        x=mult(x,x);        n>>=1;    }    return res.m[0][0];}int main(){    int n;    while(scanf("%d",&n)&&n!=-1)    {        if(n==0)        {            printf("0\n");            continue;        }        int ans=pow_mod(n-1);        printf("%d\n",ans);    }    return 0;}

这存在个公式，明白这个公式，就知道怎样写了,至于推导过程因为打出来较麻烦，以后有空时我在补充。