矩阵快速幂模板

Description

In the Fibonacci integer sequence, F_0 = 0, F_1 = 1, and F_n = F_{n-1} + F_{n-2} for n \geq 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \cdots

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 \leq n \leq 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 \times 2 matrices is given by

Also, note that raising any 2 \times 2 matrix to the 0th power gives the identity matrix:

The data used in this problem is unofficial data prepared by 695375900. So any mistake here does not imply mistake in the offcial judge data.

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <algorithm>

 5 #include<cmath>

 6 #include<sstream>

 7 #include<string>

 8 using namespace std;

 9 #define M 10000

10 struct matrix

11 {

12     int a[2][2];

13 };

14 matrix mul(matrix x,matrix y)

15 {

16     matrix temp;

17     memset(temp.a,0,sizeof(temp.a));

18     for(int i=0;i<2;i++)

19         for(int j=0;j<2;j++)

20         for(int k=0;k<2;k++)

21         temp.a[i][j]=(temp.a[i][j]+x.a[i][k]*y.a[k][j])%M;

22     return temp;

23 }

24 matrix mpow(matrix A,int n)

25 {

26     matrix B;

27     memset(B.a,0,sizeof(B.a));

28     for(int i=0;i<2;i++)

29         B.a[i][i]=1;

30     while(n>0)

31     {

32         if(n&1)

33             B=mul(B,A);

34         A=mul(A,A);

35         n>>=1;

36     }

37     return B;

38 }

39 int main()

40 {

41    matrix A;

42    int n;

43    while(~scanf("%d",&n))

44    {

45    A.a[0][0]=1;A.a[0][1]=1;

46    A.a[1][0]=1;A.a[1][1]=0;

47     if(n==-1)

48    break;

49     if(n==0)

50    {printf("0\n");

51    continue;}

52    if(n==1)

53     {printf("1\n");

54     continue;}

55 

56     A=mpow(A,n);

57    printf("%d\n",A.a[1][0]);}

58     return 0;

59 }