POJ

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16668 Accepted: 11670

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 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 of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

题意：给定一个很大的数n，求斐波那契数列中的第n个数模10000的结果。

思路：

在斐波那契数列中，有 f(n) = 1*f(n-1) + f(n-2)    f(n-1) = 1*f(n-1) + f(n-2) ，所以

所以只要求出矩阵的n-1次方是多少就可以了。

和数的快速幂相似，把数的乘法换成矩阵的乘法即可。

最后要记得特判0...

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#define max_ 200010#define inf 0x3f3f3f3f#define ll long longusing namespace std;struct mat{    ll num[2][2];};struct mat a;mat mult(mat a,mat b){    mat ans;    memset(ans.num,0,sizeof(ans.num));    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            for(int k=0;k<2;k++)            {                ans.num[i][j]+=(a.num[i][k]*b.num[k][j]);                ans.num[i][j]%=10000;            }        }    }    return ans;}int fpow(ll b){    mat ans;    ans.num[0][0]=1;    ans.num[1][1]=1;    ans.num[0][1]=0;    ans.num[1][0]=0;    mat tmp;    tmp.num[0][0]=1;    tmp.num[1][1]=0;    tmp.num[0][1]=1;    tmp.num[1][0]=1;    while(b!=0)    {        if(b&1)            ans=mult(ans,tmp);        b/=2;        tmp=mult(tmp,tmp);    }    return ans.num[0][0];}int main(int argc, char const *argv[]) {    ll n;    while(cin>>n&&n!=-1)    {        if(n==0)        printf("0\n");        else        printf("%d\n",fpow(n-1));    }    return 0;}