UVA 10689 Yet another Number Sequence

Input: standard input 
Output: standard output 
Time Limit: 3 seconds

Let’s define another number sequence, given by the following function: 
f(0) = a 
f(1) = b 
f(n) = f(n-1) + f(n-2), n > 1 
When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and b , you can get many different sequences. Given the values of a, b, you have to find the last m digits of f(n) . 
Input

The first line gives the number of test cases, which is less than 10001. Each test case consists of a single line containing the integers a b n m. The values of a and b range in [0,100], value of n ranges in [0, 1000000000] and value of m ranges in [1, 4].

Output 
For each test case, print the last m digits of f(n). However, you should NOT print any leading zero.

Sample Input Output for Sample Input 
4 
0 1 11 3 
0 1 42 4 
0 1 22 4 
0 1 21 4

89 
4296 
7711 
946 
Problem setter: Sadrul Habib Chowdhury 
Special Thanks: Derek Kisman, Member of Elite Problem Setters’ Panel

#include <cstdio>#include <cstring>const int N = 2;int MOD;struct Matrix{int ary[N][N];void init() {memset(ary, 0, sizeof(ary));}Matrix() {init();}};const Matrix operator*(const Matrix & A, const Matrix & B) {Matrix t;for (int i = 0; i < N; ++i)for (int j = 0; j < N; ++j) {for (int k = 0; k < N; ++k)t.ary[i][j] += A.ary[i][k] * B.ary[k][j];t.ary[i][j] %= MOD;}return t;}int quick_pow(int a, int b, int n) {if (n == 0) return a % MOD;if (n == 1) return b % MOD;Matrix ans, tmp;tmp.ary[0][0] = 1;ans.ary[0][0] = (a + b) % MOD;for (int i = 0; i < N; ++i) {tmp.ary[i][1 - i] = 1;ans.ary[i][1 - i] = b % MOD;}n -= 2;while (n) {if (n & 1)ans = ans * tmp;n >>= 1;tmp = tmp * tmp;}return ans.ary[0][0];}int main() {int T;scanf("%d", &T);while (T--) {int a, b, n, m;scanf("%d%d%d%d", &a, &b, &n, &m);MOD = 1;while (m--) MOD *= 10;printf("%d\n", quick_pow(a, b, n));}return 0;}