POJ 3070

Description

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

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

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Solution
将斐波那契数列用下图方法表示出来，然后取右上角即可。

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Code

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;#define maxn 3typedef long long ll;struct Mat{    int mat[maxn][maxn];//矩阵     int row, col;//矩阵行列数 };Mat mod_mul(Mat a, Mat b, int p)//矩阵乘法 {    Mat ans;    ans.row = a.row;    ans.col = b.col;    memset(ans.mat, 0, sizeof(ans.mat));    for (int i = 1;i<=ans.row;i++)        for (int j = 1;j<=ans.col;j++)            for (int k = 1;k<=a.col;k++)            {                ans.mat[i][j] += a.mat[i][k] * b.mat[k][j];                ans.mat[i][j] %= p;            }    return ans;}Mat mod_pow(Mat a, int k, int p)//矩阵快速幂 {    Mat ans;    ans.row = a.row;    ans.col = a.col;    for (int i = 1;i <= a.row;i++)        for (int j = 1;j <= a.col;j++)            ans.mat[i][j] = (i == j);    while (k)    {        if (k & 1)ans = mod_mul(ans, a, p);        a = mod_mul(a, a, p);        k >>= 1;    }    return ans;}int main(){    int n;    Mat A;    A.mat[1][1] = 1; A.mat[1][2] = 1;    A.mat[2][1] = 1; A.mat[2][2] = 0;    A.row = A.col = 2;    while (scanf("%d",&n)&&n!=-1)    {        Mat ans = mod_pow(A, n, 10000);        printf("%d\n", ans.mat[1][2]);    }    return 0;}