In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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.

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).

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

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

.

 1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 const int mod = 10000; 5 using namespace std; 6 struct mat { 7     int a[3][3]; 8     mat() { 9         memset(a, 0, sizeof(a));10     }11     mat operator *(const mat &o) const {12         mat t;13         for(int i = 1; i <= 2; i++) {14             for(int j = 1; j <= 2; j++)15                 for(int k = 1; k <= 2; k++) {16                     t.a[i][j] = (t.a[i][j] + a[i][k]*o.a[k][j])%mod;17                 }18         }19         return t;20     }21 } a, b;22 23 int main() {24     int n;25     while(scanf("%d", &n) !=EOF && n!= -1) {26         a.a[1][1] = a.a[2][2] = 1, a.a[1][2] = a.a[2][1] = 0;27         b.a[1][1] = b.a[1][2] = b.a[2][1] = 1, b.a[2][2] = 0;28         if(n == 0) printf("0\n");29         else {30             n--;31             while(n > 0) {32                 if(n&1) {33                     a = b*a;34                     n--;35                 }36                 n >>= 1;37                 b = b*b;38             }39             printf("%d\n", a.a[1][1]);40         }41     }42     return 0;43 }