Fibonacci

最新推荐文章于 2021-03-28 14:18:10 发布

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本文介绍了一种计算斐波那契数列中任意项的最后四位数字的方法，并通过矩阵快速幂算法提高了计算效率。特别适用于大数值场景，避免了直接计算导致的溢出问题。

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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

#include<cstdio>
#include<cstring>
using namespace std;
typedef struct Matrix
{
    int s[2][2];
} M;
M Multiply(M a,M b)
{
   M c;
   memset(c.s,0,sizeof(c.s));
   for(int i=0;i<2;i++)
   {
       for(int j=0;j<2;j++)
       {
           for(int k=0;k<2;k++)
           {
               c.s[i][j]+=a.s[i][k]*b.s[k][j]%10000;
           }
       }
   }
   return c;
}
M pow(M a,int n)
{
    M b;
    if(n==0)
    {
        memset(b.s,0,sizeof(b.s));
        b.s[0][0]=1;
        b.s[1][1]=1;
        return b;
    }
    else
    {
       M k=pow(a,n/2);
       if(n%2==1)
       return Multiply(Multiply(k,k),a);
       else return Multiply(k,k);
    }
}
int main()
{
    int n;
    while(scanf("%d",&n),n!=-1)
    {
        M a;
        a.s[0][0]=1;
        a.s[0][1]=1;
        a.s[1][0]=1;
        a.s[1][1]=0;
        M b=pow(a,n);
        printf("%d\n",b.s[0][1]%10000);
    }
    return 0;
}

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