集训队专题（2）1001 Fibonacci

Fibonacci

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 53   Accepted Submission(s) : 41

Problem 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

Source

0
34
626
6875      

PKU

最基础的矩阵快速幂用法，题目已经给出了具体的公式那就不在赘述，代码：

#include <cstdio>
#include <iostream>
#define MOD 10000
using namespace std;
struct matrix
{
  int m[2][2];
}ans,base;
matrix multi(matrix a,matrix b)
{
  matrix tmp;
  for(int i=0; i<2; i++)
  {
    for(int j=0; j<2; j++)
    {
      tmp.m[i][j] = 0;
      for(int k=0; k<2; k++)
        tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k]*b.m[k][j]) % MOD;
    }
  }
  return tmp;
}
int fast_mod(int n)
{
  base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
  base.m[1][1] = 0;
  ans.m[0][0] = ans.m[1][1] = 1;
  ans.m[0][1] = ans.m[1][0] = 0;
  while(n)
  {
    if(n&1) ans = multi(ans,base);
    base = multi(base,base);
    n >>= 1;
  }
  return ans.m[0][1];
}
int main()
{
  int n;
  while(scanf("%d",&n)&&n!=-1)
  {
    printf("%d\n",fast_mod(n));
  }
  return 0;
}