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179 lines
4.4 KiB
Python
Executable file
179 lines
4.4 KiB
Python
Executable file
"""#
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### 谜题描述
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Ayoub had an array a of integers of size n and this array had two interesting properties:
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* All the integers in the array were between l and r (inclusive).
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* The sum of all the elements was divisible by 3.
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Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array.
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Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.
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Input
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The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.
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Output
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Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.
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Examples
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Input
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2 1 3
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Output
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3
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Input
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3 2 2
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Output
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1
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Input
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9 9 99
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Output
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711426616
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Note
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In the first example, the possible arrays are : [1,2], [2,1], [3, 3].
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In the second example, the only possible array is [2, 2, 2].
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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from sys import stdin
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from copy import deepcopy
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n, l, r = map(int, stdin.readline().split())
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l -= 1
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div1, div2, mod = l // 3, r // 3, 1000000007
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all = [div2, div2 + 1 if r % 3 else div2, div2 + (min(r % 3, 2) // 2)]
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minus = [div1, div1 + 1 if l % 3 else div1, div1 + (min(l % 3, 2) // 2)]
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all = [all[i] - minus[i] for i in range(3)]
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mem, p = [deepcopy(all), [0, 0, 0]], 0
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for i in range(1, n):
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p ^= 1
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for j in range(1, 4):
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for k in range(1, 4):
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tem = (j + k) % 3
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mem[p][tem] = mem[p][tem] % mod + (mem[p ^ 1][j % 3] * all[k % 3]) % mod
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for j in range(3):
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mem[p ^ 1][j] = 0
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print(mem[p][0] % mod)
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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import re
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from bootcamp import Basebootcamp
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def compute_answer(n, l, r):
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mod = 10**9 + 7
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def count_mod(low, high, m):
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remainder = low % 3
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if remainder <= m:
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first = low + (m - remainder)
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else:
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first = low + (3 - remainder + m)
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if first > high:
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return 0
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last = high - ((high - m) % 3)
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return ((last - first) // 3) + 1
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count0 = count_mod(l, r, 0)
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count1 = count_mod(l, r, 1)
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count2 = count_mod(l, r, 2)
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counts = [count0, count1, count2]
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# Dynamic programming approach
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dp_prev = counts.copy()
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for _ in range(n - 1):
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dp_next = [0] * 3
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for prev_mod in range(3):
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for curr_mod in range(3):
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new_mod = (prev_mod + curr_mod) % 3
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dp_next[new_mod] = (dp_next[new_mod] + dp_prev[prev_mod] * counts[curr_mod]) % mod
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dp_prev = dp_next
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return dp_prev[0] % mod
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class Cayoubandlostarraybootcamp(Basebootcamp):
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def __init__(self, n_min=1, n_max=20, l_min=1, r_max=10**9):
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self.n_min = n_min
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self.n_max = n_max
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self.l_min = l_min
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self.r_max = r_max
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def case_generator(self):
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n = random.randint(self.n_min, self.n_max)
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l = random.randint(self.l_min, self.r_max)
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r = random.randint(l, self.r_max)
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correct_answer = compute_answer(n, l, r)
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return {
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'n': n,
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'l': l,
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'r': r,
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'correct_answer': correct_answer
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}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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l = question_case['l']
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r = question_case['r']
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problem = f"""Given three integers n, l, and r, calculate the number of arrays of length n where each element is between l and r (inclusive) and the total sum is divisible by 3. Return the result modulo 10^9+7.
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Input:
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n = {n}, l = {l}, r = {r}
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Put your final answer within [answer] tags like [answer]123[/answer]."""
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return problem
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@staticmethod
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def extract_output(output):
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pattern = r'\[answer\](.*?)\[/answer\]'
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matches = re.findall(pattern, output, re.DOTALL)
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if not matches:
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return None
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last_match = matches[-1].strip()
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try:
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return int(last_match)
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except ValueError:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['correct_answer']
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