poj 3070 Fibonacci

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11123 Accepted: 7913

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

Hint

题意：给你一个求解第N个斐波那契数的公式。 让你求出Fn % 10000。

#include <iostream>#include <string.h>using namespace std;struct matrix{    long long e[2][2];};int n,mod=10000;matrix matrix_mul(matrix a,matrix b){    matrix sum;    memset(sum.e,0,sizeof(sum.e));    for(int i=0;i<2;++i)        for(int j=0;j<2;++j)        for(int k=0;k<2;++k)        if(a.e[i][k]&&b.e[k][j])        sum.e[i][j]=(sum.e[i][j]+a.e[i][k]*b.e[k][j])%mod;        return sum;}long long matrix_pow(){    matrix s,ee;    ee.e[0][0]=ee.e[1][1]=1;    ee.e[0][1]=ee.e[1][0]=0;    s.e[0][0]=s.e[0][1]=s.e[1][0]=1;    s.e[1][1]=0;    while(n)    {        if(n&1)            ee=matrix_mul(ee,s);        n>>=1;        s=matrix_mul(s,s);    }    return ee.e[1][0];//为什么矩阵中处于【1】【0】这个位置的值就是所要的结果还不理解}int main(){    while(cin>>n,n!=-1)    {        if(n==0||n==1)            cout<<n<<endl;        else            cout<<matrix_pow()<<endl;    }    return 0;}