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

0
34
626
6875

Solution

Code

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48  #include #include #include #include using namespace std; #define maxn 3 typedef long long ll; struct Mat { int mat[maxn][maxn];//矩阵 int row, col;//矩阵行列数 }; Mat mod_mul(Mat a, Mat b, int p)//矩阵乘法 { Mat ans; ans.row = a.row; ans.col = b.col; memset(ans.mat, 0, sizeof(ans.mat)); for (int i = 1;ifor (int j = 1;jfor (int k = 1;kreturn ans; } Mat mod_pow(Mat a, int k, int p)//矩阵快速幂 { Mat ans; ans.row = a.row; ans.col = a.col; for (int i = 1;i for (int j = 1;j while (k) { if (k & 1)ans = mod_mul(ans, a, p); a = mod_mul(a, a, p); k >>= 1; } return ans; } int main() { int n; Mat A; A.mat[1][1] = 1; A.mat[1][2] = 1; A.mat[2][1] = 1; A.mat[2][2] = 0; A.row = A.col = 2; while (scanf("%d",&n)&&n!=-1) { Mat ans = mod_pow(A, n, 10000); printf("%d\n", ans.mat[1][2]); } return 0; }