POJ 3070：Fibonacci

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15071 Accepted: 10581

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

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 <iostream>#include <stdio.h>#include <math.h>#include <string.h>long long n;  ///问题规模const int N = 5;int temp[N][N];int res[N][N];void Mul(int a[][N],int b[][N]){    memset(temp,0,sizeof(temp));    for(int i = 0; i < 2; i++)  ///i行        for(int j = 0; j < 2; j++)  ///j列            for(int k = 0; k < 2; k++)                temp[i][j] = (temp[i][j]+a[i][k]*b[k][j])%10000;    for(int i = 0; i < 2; i++)        for(int j = 0; j < 2; j++)            a[i][j] = temp[i][j];}void Solve(int a[][N],int n)  ///n是求的幂次{    memset(res,0,sizeof(res));    for(int i = 0; i < 2; i++)        res[i][i] = 1;    while(n)    {        if(n&1)            Mul(res,a);        Mul(a,a);        n>>=1;    }}int main(){    int T;    while(~scanf("%d",&T))    {        int a[N][N];        a[0][0] = 1;        a[0][1] = 1;        a[1][0] = 1;        a[1][1] = 0;        if(T == -1) break;        if(T == 0 || T == 1) printf("%d\n",T);        else        {            Solve(a,T);            printf("%d\n",res[0][1]);        }    }    return 0;}