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288 lines
7.3 KiB
Python
Executable file
288 lines
7.3 KiB
Python
Executable file
"""#
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### 谜题描述
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The sequence of integer pairs (a1, b1), (a2, b2), ..., (ak, bk) is beautiful, if the following statements are fulfilled:
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* 1 ≤ a1 ≤ b1 < a2 ≤ b2 < ... < ak ≤ bk ≤ n, where n is a given positive integer;
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* all numbers b1 - a1, b2 - a2, ..., bk - ak are distinct.
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For the given number n find the number of beautiful sequences of length k. As the answer can be rather large, print the remainder after dividing it by 1000000007 (109 + 7).
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Input
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The first line contains integer t (1 ≤ t ≤ 2·105) — the number of the test data.
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Each of the next t lines contains two integers n and k (1 ≤ k ≤ n ≤ 1000).
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Output
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For each test from the input print the answer to the problem modulo 1000000007 (109 + 7). Print the answers to the tests in the order in which the tests are given in the input.
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Examples
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Input
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6
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1 1
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2 1
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2 2
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3 1
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3 2
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3 3
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Output
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1
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3
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0
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6
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2
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0
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Note
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In the first test sample there is exactly one beautiful sequence: (1, 1).
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In the second test sample, the following sequences are beautiful:
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* (1, 1);
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* (1, 2);
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* (2, 2).
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In the fourth test sample, the following sequences are beautiful:
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* (1, 1);
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* (1, 2);
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* (1, 3);
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* (2, 2);
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* (2, 3);
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* (3, 3).
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In the fifth test sample, the following sequences are beautiful:
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* (1, 1), (2, 3);
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* (1, 2), (3, 3).
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In the third and sixth samples, there are no beautiful sequences.
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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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#include <bits/stdc++.h>
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using namespace std;
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const double pi = acos(-1);
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const double eps = 1e-10;
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const int inf = 0x3f3f3f3f;
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const long long infLL = 0x3f3f3f3f3f3f3f3fLL;
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const int maxn = 2000 + 5;
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const long long mod = 1000000007;
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long long fac[maxn], ifac[maxn];
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void gcd(long long a, long long b, long long &d, long long &x0, long long &y0) {
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if (!b) {
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d = a;
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x0 = 1;
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y0 = 0;
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} else {
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gcd(b, a % b, d, y0, x0);
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y0 -= x0 * (a / b);
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}
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}
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long long inv(long long a, long long m = mod) {
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long long d, x, y;
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gcd(a, m, d, x, y);
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return d == 1 ? (x + m) % m : -1;
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}
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void mk_fac() {
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fac[0] = 1;
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for (int i = 1; i < maxn; ++i) fac[i] = fac[i - 1] * i % mod;
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for (int i = 0; i < maxn; ++i) ifac[i] = inv(fac[i]);
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}
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long long A(int n, int m) {
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if (!fac[0]) mk_fac();
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return fac[n] * ifac[n - m] % mod;
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}
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long long C(int n, int m) {
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if (!fac[0]) mk_fac();
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return fac[n] * ifac[n - m] % mod * ifac[m] % mod;
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}
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long long s[maxn];
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long long f[maxn][maxn];
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void init() {
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s[0] = 0;
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for (int i = 1; i < maxn; ++i) s[i] = s[i - 1] + i;
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for (int i = 1; i < maxn; ++i) f[i][1] = 1;
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for (int j = 2; j < maxn; ++j) {
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if (s[j] >= maxn) break;
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f[s[j]][j] = fac[j];
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for (int i = s[j] + 1; i < maxn; ++i) {
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f[i][j] += f[i - j][j] + f[i - j][j - 1] * j % mod;
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f[i][j] %= mod;
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}
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}
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}
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int main() {
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ios::sync_with_stdio(false);
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mk_fac();
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init();
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int T;
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cin >> T;
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while (T--) {
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int n, k;
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cin >> n >> k;
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--n;
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long long res = 0;
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for (int i = s[k - 1]; i <= n; ++i) {
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int t = n - i - (k - 1);
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if (t < 0) break;
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res += f[i + k][k] * C(k + t, t) % mod;
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res %= mod;
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}
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cout << res << endl;
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}
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return 0;
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}
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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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class Dbeautifulpairsofnumbersbootcamp(Basebootcamp):
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mod = 10**9 + 7
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maxn = 2005
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fac = []
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ifac = []
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s = []
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f = []
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initialized = False
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def __init__(self, max_n=1000, min_n=1, **kwargs):
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super().__init__(**kwargs)
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self.max_n = max_n
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self.min_n = min_n
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self.initialize_data()
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@classmethod
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def initialize_data(cls):
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if cls.initialized:
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return
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# Precompute factorial and inverse factorial arrays
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cls.fac = [1] * cls.maxn
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for i in range(1, cls.maxn):
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cls.fac[i] = cls.fac[i-1] * i % cls.mod
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cls.ifac = [1] * cls.maxn
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cls.ifac[cls.maxn - 1] = pow(cls.fac[cls.maxn - 1], cls.mod - 2, cls.mod)
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for i in range(cls.maxn - 2, -1, -1):
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cls.ifac[i] = cls.ifac[i + 1] * (i + 1) % cls.mod
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# Precompute s array
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cls.s = [0] * cls.maxn
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for i in range(1, cls.maxn):
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cls.s[i] = cls.s[i-1] + i
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# Initialize f array using dynamic programming
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cls.f = [[0] * cls.maxn for _ in range(cls.maxn)]
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for i in range(1, cls.maxn):
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cls.f[i][1] = 1
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for j in range(2, cls.maxn):
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if cls.s[j] >= cls.maxn:
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break
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if cls.s[j] < cls.maxn:
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cls.f[cls.s[j]][j] = cls.fac[j] % cls.mod
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for i in range(cls.s[j] + 1, cls.maxn):
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prev_i = i - j
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if prev_i >= 0:
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term1 = cls.f[prev_i][j]
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term2 = (cls.f[prev_i][j-1] * j) % cls.mod
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cls.f[i][j] = (term1 + term2) % cls.mod
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cls.initialized = True
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@classmethod
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def compute_answer(cls, n, k):
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if k < 1 or k > n:
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return 0
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new_n = n - 1
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res = 0
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s_k_1 = cls.s[k-1]
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for i in range(s_k_1, new_n + 1):
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t = new_n - i - (k - 1)
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if t < 0:
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break
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comb = cls.C(k + t, t)
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if (i + k) >= cls.maxn or k >= cls.maxn:
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f_val = 0
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else:
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f_val = cls.f[i + k][k]
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res = (res + f_val * comb) % cls.mod
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return res
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@classmethod
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def C(cls, n, m):
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if m < 0 or m > n:
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return 0
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return cls.fac[n] * cls.ifac[m] % cls.mod * cls.ifac[n - m] % cls.mod
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def case_generator(self):
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n = random.randint(self.min_n, self.max_n)
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k = random.randint(1, n)
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correct_answer = self.compute_answer(n, k)
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return {
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'n': n,
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'k': k,
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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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k = question_case['k']
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prompt = f"""你是一个算法竞赛选手，现在需要解决一个数学谜题。请仔细阅读以下问题描述，并输出你的答案。
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问题描述:
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给定一个正整数n和k，计算满足条件的“美丽序列”的数量。答案需要对1e9+7取模。
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美丽序列的定义:
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- 序列由k个整数对组成：(a1, b1), (a2, b2), ..., (ak, bk)。
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- 满足以下两个条件：
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1. 所有整数对严格递增且互不重叠，即1 ≤ a1 ≤ b1 < a2 ≤ b2 < ... < ak ≤ bk ≤ n。
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2. 每个整数对的差（即bi - ai）互不相同。
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输入要求:
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- n的值为{n}，k的值为{k}。
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输出要求:
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- 输出满足条件的美丽序列的数量模1000000007的结果。
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请将最终答案放在[answer]和[/answer]的标签之间。例如：[answer]42[/answer]。"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', 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:
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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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