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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<cmath>
#include<sstream>
#include<string>
using namespace std;
#define M 10000
struct matrix
{
    int a[2][2];
};
matrix mul(matrix x,matrix y)
{
    matrix temp;
    memset(temp.a,0,sizeof(temp.a));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
        for(int k=0;k<2;k++)
        temp.a[i][j]=(temp.a[i][j]+x.a[i][k]*y.a[k][j])%M;
    return temp;
}
matrix mpow(matrix A,int n)
{
    matrix B;
    memset(B.a,0,sizeof(B.a));
    for(int i=0;i<2;i++)
        B.a[i][i]=1;
    while(n>0)
    {
        if(n&1)
            B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    }
    return B;
}
int main()
{
   matrix A;
   int n;
   while(~scanf("%d",&n))
   {
   A.a[0][0]=1;A.a[0][1]=1;
   A.a[1][0]=1;A.a[1][1]=0;
    if(n==-1)
   break;
    if(n==0)
   {printf("0\n");
   continue;}
   if(n==1)
    {printf("1\n");
    continue;}

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