UAa11582--Colossal Fibonacci Numbers!(快速冪)

Do more with less

Discussion

The i’th Fibonacci number f(i) is
recursively defined in the following
way:
•f(0)=0 andf(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

題意

輸入三個非負整數a,b,n；計算f(ab)modn

思路

劉大爺的數316頁，自己對這道題沒想法

代碼

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn = 1000 + 5;
typedef unsigned long long ULL;

int f[maxn][maxn*6], period[maxn];

ULL pow_mod(ULL a, ULL b, int n) {
  if(!b) return 1;
  ULL k = pow_mod(a, b/2, n);
  k = k * k % n;
  if(b % 2) k = k * a % n;
  return k;
}

int solve(ULL a, ULL b, int n) {
  if(a == 0 || n == 1) return 0; // attention!
  int p = pow_mod(a % period[n], b, period[n]);
  return f[n][p];
}

int main() {
  for(int n = 2; n <= 1000; n++) {
    f[n][0] = 0; f[n][1] = 1;
    for(int i = 2; ; i++) {
      f[n][i] = (f[n][i-1] + f[n][i-2]) % n;
      if(f[n][i-1] == 0 && f[n][i] == 1) {
        period[n] = i - 1;
        break;
      }
    }
  }
  ULL a, b;
  int n, T;
  cin >> T;
  while(T--) {
    cin >> a >> b >> n;
    cout << solve(a, b, n) << "\n";
  }
  return 0;
}