PTA Logs Stacking (25 points) (Fibonacci sequence properties)

Rapid doubling

Reference Code

pair<int, int> fib(int n) {
    
    
  if (n == 0) return {
    
    0, 1};
  auto p = fib(n >> 1);
  int c = p.first * (2 * p.second - p.first);
  int d = p.first * p.first + p.second * p.second;
  if (n & 1)
    return {
    
    d, c + d};
  else
    return {
    
    c, d};
}

The code is the source of the article: https://oi-wiki.org/math/fibonacci/

Periodicity in the modular sense

topic

7-13 Logs Stacking (25分)

Daxinganling produces a lot of timber. Before loading onto trains, the timber jacks will place the logs to some place within no more than two logs’ height first. Looking from the sideway, the figure of a logs stack is as follows:

We have known that the number of logs in each layer is fewer than the lower layer for at least one log, and that in each layer the logs are connected in a line.

In the figure above, there are 4 logs in the bottom layer of the stack, and 3 logs in the top layer of the stack. So the picture shows one kind of figures with 7 logs accumulated together.

Now, given the total number of logs, Little Gyro wants to know how many possible figures there may be.

Input Specification:

There are multiple test cases. The first line of the input contains an integer T (1 ≤ T ≤ 10
​5
​​ ), indicating the number of test cases. For each test case:

Each line contains one integer n (1 ≤ n ≤ 2×10
​9
​​ ), indicating the total number of logs.

Output Specification:

For each test case in the input, you should output the corresponding number of possible figures.

Because the number may be very large, just output the number mod 10000.

Sample Input:

5
1
3
5
7
2000000000

Sample Output:

1
2
5
13
3125

Hint:

In the third sample, you can accumulate 5 logs within such following ways:

First, you can put all the 5 logs at the ground floor.

Then, you can put 4 logs at the bottom layer, and there are 3 positions for the last log to pile up.

After that, you can also put 3 logs at the bottom layer, while the other 2 logs at the top layer.

Above all, you can get 1 + 3 + 1 = 5 figures in order to accumulate 5 logs.

Problem solving

Use the above properties to list the recurrence formula, and find the rule by typing the table. The period T is 75000

Code

#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
const int maxn = 1e4 + 2;
const int p = 10000;
int Fib[p];

// 奇数 F(2k + 1) = F(k + 1) ^ 2 + F(k) ^ 2;
// 偶数 F(2k) = F(k)(2F(k + 1) - F(k));
// 该题周期性, 周期为75000

int fib(int n) {
    
    
    // if (n == 2 || n == 1) return 1;
    if (n < maxn) return Fib[n];
    int a = fib(n / 2) % p;      // F(k)
    int b = fib(n / 2 + 1) % p;  // F(k + 1)
    if (n & 1) {
    
                     //为奇数
        return (a * a % p + b * b % p) % p;
    } else {
    
      //为偶数
        return a * (2 * b - a + p) % p;
    }
}

int main() {
    
    
    Fib[1] = 1;
    for (int i = 2; i < maxn; i++) {
    
    
        Fib[i] = (Fib[i - 1] + Fib[i - 2]) % p;
    }

    int T, n;
    cin >> T;
    while (T--) {
    
    
        scanf("%d", &n);
        cout << fib(n % 75000) << endl;
    }

    system("pause");
    return 0;
}