nyoj-148-fibonacci数列(二)

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描述

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.

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:

.

输入

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.

输出

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).

样例输入

0
9
1000000000
-1

样例输出

0
34
6875

矩阵的快速幂

#include<stdio.h>
#include<string.h>
int main()
{int n,i,j,k;while(scanf("%d",&n)&&n!=-1){int a[2][2]={1,0,1,0};int b[2][2]={1,1,1,0};int c[2][2];while(n){if(n&1){memset(c,0,sizeof(c));for(i=0;i<2;i++)for(j=0;j<2;j++)for(k=0;k<2;k++)c[i][k]+=(a[i][j]*b[j][k])%10000;for(i=0;i<2;i++)for(j=0;j<2;j++)a[i][j]=c[i][j];}memset(c,0,sizeof(c));for(i=0;i<2;i++)for(j=0;j<2;j++)for(k=0;k<2;k++)c[i][k]+=(b[i][j]*b[j][k])%10000;for(i=0;i<2;i++)for(j=0;j<2;j++)b[i][j]=c[i][j];n>>=1;}printf("%d\n",a[1][1]%10000);}return 0;
}

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