Colossal Fibonacci Numbers!
Oooh...pretty
The i’th Fibonacci number f(i) is recursively
defined in the following way:
• f(0) = 0 and f(1) = 1
• f(i + 2) = f(i + 1) + f(i) for every
i 0
Your task is to compute some values
of this sequence.
Input
Input begins with an integer t 10; 000,
the number of test cases. Each test case
consists of three integers a, b, n where
0 a; b < 264 (a and b will not both be
zero) and 1 n 1000.
Output
For each test case, output a single line containing the remainder of f(ab) upon division by n.
Sample Input
31
1 2
2 3 1000
18446744073709551615 18446744073709551615 1000
Sample Output
1
21
250
题意：
给出64位整数a、b以及不超过1000的正整数n，求斐波那契数列第a ^ b项模n的结果。
输入：
情况数T，之后T行每行a、b、n。
输出：
斐波那契数列第a ^ b项模n的结果。
分析：
由于斐波那契数列的每一项都是由前两项相加得来，并且每一项都对n取模，所有每一项的情况一共有n种，而相邻两项若组成有序数对，则不同的数对的情况也只有n ^ 2种，所以只需要计算n * n项就可以找到数列规律。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 using namespace std;
5 typedef unsigned long long ull;
6 const int maxn = 1000;
7 ull F[maxn * maxn + 10];
8 int power(ull a,ull b,int t){
9 ull res = 1;
10 while(b){
11 if(b & 1) res = (a * res) % t;
12 a = (a * a) % t;
13 b >>= 1;
14 }
15 return res;
16 }
17 int main()
18 {
19 int T;
20 cin >> T;
21 while(T--)
22 {
23 ull a,b,n,l;
24 int flag = 1;
25 cin >> a >> b >> n;
26 if(n == 1) {printf("0\n"); continue;}
27 F[0] = 0;
28 F[1] = 1;
29 for(ull i = 2 ; ; i++){
30 F[i] = (F[i - 1] % n + F[i - 2] % n) % n;
31 if(F[i] == F[1] && F[i - 1] == F[0]){
32 l = i - 1;
33 break;
34 }
35 }
36 cout << F[power(a % l, b, l)] << endl;
37 }
38 return 0;
39 }