UVa11582

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。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef unsigned long long ull;
const int maxn = 1000;
ull F[maxn * maxn + 10];
int power(ull a,ull b,int t){
    ull res = 1;
    while(b){
        if(b & 1) res = (a * res) % t;
        a = (a * a) % t;
        b >>= 1;
    }
    return res;
}
int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        ull a,b,n,l;
        int flag = 1;
        cin >> a >> b >> n;
        if(n == 1) {printf("0
"); continue;}
        F[0] = 0;
        F[1] = 1;
        for(ull i = 2 ; ; i++){
            F[i] = (F[i - 1] % n + F[i - 2] % n) % n;
            if(F[i] == F[1] && F[i - 1] == F[0]){
                l = i - 1;
                break;
            }
        }
        cout << F[power(a % l, b, l)] << endl;
    }
    return 0;
}