UVA 11582 Colossal Fibonacci Numbers! 快速幂

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2629

题目大意：菲波那切数列求f(a^b) % n，其中 a,b < 2⁶⁴ ，n < 1001.

解题思路：打表找规律。然后求出f[i]%n的周期之后快速幂取模。

坑点：unsigned long long！！

代码：

 1 const int maxn = 1e6 + 10;
 2 int f[maxn] = {0, 1};
 3 int n, m;
 4 unsigned long long a, b;
 5 
 6 unsigned long long  pow_mod(unsigned long long x, unsigned long long y){
 7     if(y == 0) return 1;
 8     ll tmp = pow_mod(x, y / 2) % m;
 9     ll ans = tmp * tmp % m;
10     if(y & 1) ans = x % m * ans;
11     return ans % m;
12 }
13 void solve(){
14     f[1] = 1 % n; 
15     for(int i = 0; i <= n * n + 10; i++){
16         if(i == 0 || i == 1) continue;
17         else{
18             f[i] = f[i - 1] + f[i - 2];
19             f[i] %= n;
20             if(f[i] == f[1] && f[i - 1] == f[0]) {
21                 m = i - 1;
22                 break;
23             }
24         }
25     }
26     int x = pow_mod(a, b);
27     printf("%d
", f[x]);
28 }
29 
30 int main(){
31     int t;
32     scanf("%d", &t);
33     while(t--){
34         cin >> a >> b;
35         cin >> n;
36         solve();
37     }
38 }

题目：

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 < 2 64 (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(a b ) upon division by n. Sample Input 3 1 1 2 2 3 1000 18446744073709551615 18446744073709551615 1000 Sample Output 1 21 250