poj 3070 Fibonacci 矩阵相乘

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7715		Accepted: 5474

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

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

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:

.

Source

Stanford Local 2006

 

想学高斯消元，先学矩阵。

二分思想，在A^B mod m中已经体现..

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 
 struct node
 {
     int a[][];
 }now,cur;
 
 void make_first()
 {
     now.a[][]=;
     now.a[][]=;
     now.a[][]=;
     now.a[][]=;
 
     cur.a[][]=;
     cur.a[][]=;
     cur.a[][]=;
     cur.a[][]=;
 }
 
 struct node make_cheng(node  a,node b) //发现结构体忘记了。
 {
     struct node f;
     int i,k,j;
     memset(f.a,,sizeof(f.a));
     for(i=;i<=;i++)
     for(k=;k<=;k++)
     if(a.a[i][k])
     {
         for(j=;j<=;j++)
         if(b.a[k][j])
         {
             f.a[i][j]+=a.a[i][k]*b.a[k][j];
             if(f.a[i][j]>=)
             f.a[i][j]%=;
         }
     }
     return f;
 }
 
 void make_EF(int n)
 {
     make_first();
     while(n)
     {
         if(n&)
         {
             now=make_cheng(now,cur);//！！！
         }
         n=n/;
         cur=make_cheng(cur,cur);
     }
     printf("%d\n",now.a[][]);
 }
 
 int main()
 {
     int n;
     while(scanf("%d",&n)>)
     {
         if(n==-)break;
         if(n==){printf("0\n");continue;}
         if(n==||n==){printf("1\n");continue;}
         make_EF(n-);
     }
     return ;
 }