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319 lines
9.3 KiB
Python
Executable file
319 lines
9.3 KiB
Python
Executable file
"""#
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### 谜题描述
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In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
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F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2).
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DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: a1, a2, ..., an. Moreover, there are m queries, each query has one of the two types:
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1. Format of the query \"1 l r\". In reply to the query, you need to add Fi - l + 1 to each element ai, where l ≤ i ≤ r.
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2. Format of the query \"2 l r\". In reply to the query you should output the value of <image> modulo 1000000009 (109 + 9).
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Help DZY reply to all the queries.
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Input
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The first line of the input contains two integers n and m (1 ≤ n, m ≤ 300000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — initial array a.
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Then, m lines follow. A single line describes a single query in the format given in the statement. It is guaranteed that for each query inequality 1 ≤ l ≤ r ≤ n holds.
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Output
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For each query of the second type, print the value of the sum on a single line.
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Examples
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Input
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4 4
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1 2 3 4
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1 1 4
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2 1 4
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1 2 4
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2 1 3
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Output
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17
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12
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Note
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After the first query, a = [2, 3, 5, 7].
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For the second query, sum = 2 + 3 + 5 + 7 = 17.
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After the third query, a = [2, 4, 6, 9].
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For the fourth query, sum = 2 + 4 + 6 = 12.
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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 int N = 3e5 + 10;
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const long long MOD = 1e9 + 9;
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long long sum[N << 2], lazy1[N << 2], lazy2[N << 2], F[N];
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inline void read(long long &x) {
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int f = 1;
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x = 0;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = x * 10 + ch - '0';
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ch = getchar();
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}
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x = x * f;
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}
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struct Matrix {
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long long v[10][10];
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int lena, lenb;
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void init1(int a, int b) {
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v[0][0] = 1;
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v[0][1] = 1;
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v[1][0] = 1;
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v[1][1] = 0;
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lena = a, lenb = b;
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}
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void init2(int a, int b) {
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v[0][0] = 1;
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v[0][1] = 0;
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v[1][0] = 0;
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v[1][1] = 1;
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lena = a, lenb = b;
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}
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void init3(long long f1, long long f2, int a, int b) {
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v[0][0] = f2;
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v[0][1] = 0;
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v[1][0] = f1;
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v[1][1] = 0;
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lena = a, lenb = b;
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}
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};
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Matrix Mul(Matrix a, Matrix b) {
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Matrix ans;
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ans.lena = a.lena, ans.lenb = b.lenb;
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for (int i = 0; i < a.lena; i++) {
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for (int j = 0; j < b.lenb; j++) {
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ans.v[i][j] = 0;
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for (int k = 0; k < a.lenb; k++)
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ans.v[i][j] = (ans.v[i][j] + a.v[i][k] * b.v[k][j]) % MOD;
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}
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}
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return ans;
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}
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void print(Matrix a) {
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for (int i = 0; i < a.lena; i++) {
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for (int j = 0; j < a.lenb; j++) {
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cout << a.v[i][j] << ' ';
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}
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cout << endl;
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}
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}
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pair<long long, long long> get_f(long long f1, long long f2, int n) {
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pair<long long, long long> ans =
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pair<long long, long long>(f2, (f1 + f2) % MOD);
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if (n < 3) return ans;
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ans = pair<long long, long long>((f1 * F[n - 2] + f2 * F[n - 1]) % MOD,
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(f1 * F[n - 1] + f2 * F[n]) % MOD);
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return ans;
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}
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void modify(int rt, int l, int r, long long f1, long long f2) {
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lazy1[rt] = (lazy1[rt] + f1) % MOD;
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lazy2[rt] = (lazy2[rt] + f2) % MOD;
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sum[rt] = (sum[rt] + (get_f(f1, f2, r - l + 2).second - f2) + MOD) % MOD;
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}
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void pushDown(int rt, int l, int mid, int r) {
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if (lazy1[rt] && lazy2[rt]) {
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modify(rt << 1, l, mid, lazy1[rt], lazy2[rt]);
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pair<long long, long long> P = get_f(lazy1[rt], lazy2[rt], mid - l + 2);
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modify(rt << 1 | 1, mid + 1, r, P.first, P.second);
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lazy1[rt] = lazy2[rt] = 0;
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}
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}
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void Build(int rt, int l, int r) {
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lazy1[rt] = lazy2[rt] = sum[rt] = 0;
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if (l == r) {
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read(sum[rt]);
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return;
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}
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int mid = (l + r) >> 1;
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Build(rt << 1, l, mid);
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Build(rt << 1 | 1, mid + 1, r);
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sum[rt] = (sum[rt << 1] + sum[rt << 1 | 1]) % MOD;
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}
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void Update(int rt, int l, int r, int x, int y, long long f1, long long f2) {
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if (l == x && r == y) {
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modify(rt, l, r, f1, f2);
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return;
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}
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int mid = (l + r) >> 1;
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pushDown(rt, l, mid, r);
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if (y <= mid)
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Update(rt << 1, l, mid, x, y, f1, f2);
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else if (x > mid)
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Update(rt << 1 | 1, mid + 1, r, x, y, f1, f2);
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else {
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pair<long long, long long> P = get_f(f1, f2, mid - x + 2);
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Update(rt << 1, l, mid, x, mid, f1, f2),
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Update(rt << 1 | 1, mid + 1, r, mid + 1, y, P.first, P.second);
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}
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sum[rt] = (sum[rt << 1] + sum[rt << 1 | 1]) % MOD;
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}
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long long Query(int rt, int l, int r, int x, int y) {
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if (l == x && r == y) {
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return sum[rt];
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}
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int mid = (l + r) >> 1;
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pushDown(rt, l, mid, r);
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if (y <= mid)
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return Query(rt << 1, l, mid, x, y);
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else if (x > mid)
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return Query(rt << 1 | 1, mid + 1, r, x, y);
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else
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return (Query(rt << 1, l, mid, x, mid) +
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Query(rt << 1 | 1, mid + 1, r, mid + 1, y)) %
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MOD;
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}
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int main() {
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F[0] = 0, F[1] = 1;
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for (int i = 2; i < N; i++) F[i] = (F[i - 1] + F[i - 2]) % MOD;
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long long n, m, opt, l, r;
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read(n);
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read(m);
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Build(1, 1, n);
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for (int i = 1; i <= m; i++) {
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read(opt);
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read(l);
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read(r);
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if (opt == 1)
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Update(1, 1, n, l, r, 1, 1);
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else
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printf(\"%I64d\n\", Query(1, 1, n, l, r));
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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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from bootcamp import Basebootcamp
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import random
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MOD = 10**9 + 9
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class Cdzylovesfibonaccinumbersbootcamp(Basebootcamp):
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def __init__(self, **params):
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"""
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参数设置：
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n: 数组长度（默认随机5-10）
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m: 查询数量（默认随机5-10）
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max_val: 初始数组最大值（默认1e9）
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"""
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self.n = params.get('n', random.randint(5, 10))
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self.m = params.get('m', random.randint(5, 10))
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self.max_val = params.get('max_val', 10**9)
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self.fib_cache = {}
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self._precompute_fib(10**6) # 预生成足够大的斐波那契数列
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def _precompute_fib(self, size):
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"""预生成斐波那契数列备用"""
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self.fib_cache[1] = 1
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self.fib_cache[2] = 1
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for i in range(3, size+1):
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self.fib_cache[i] = (self.fib_cache[i-1] + self.fib_cache[i-2]) % MOD
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def case_generator(self):
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"""动态生成测试用例并计算正确答案"""
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n, m = self.n, self.m
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# 生成初始数组
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arr = [random.randint(1, self.max_val) for _ in range(n)]
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queries = []
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answers = []
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sim_arr = arr.copy()
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# 生成查询序列（保证至少2个类型2查询）
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type2_count = 0
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for _ in range(m):
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# 确保最后两个查询是类型2（保证有答案）
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if _ >= m-2 and type2_count < 2:
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query_type = 2
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else:
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query_type = random.choices([1, 2], weights=[0.7, 0.3], k=1)[0]
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l = random.randint(1, n)
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r = random.randint(l, n)
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# 处理查询
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if query_type == 1:
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# 应用斐波那契叠加
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for i in range(l-1, r):
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fib_num = (i - (l-1)) + 1 # F_{i-l+1} 的正确计算
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sim_arr[i] = (sim_arr[i] + self.fib_cache[fib_num]) % MOD
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else:
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# 计算区间和
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current_sum = sum(sim_arr[l-1:r]) % MOD
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answers.append(current_sum)
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type2_count += 1
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queries.append({"type": query_type, "l": l, "r": r})
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return {
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"n": n,
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"m": m,
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"array": arr,
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"queries": queries,
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"answers": answers
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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input_lines = [
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f"{question_case['n']} {question_case['m']}",
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' '.join(map(str, question_case['array']))
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]
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for q in question_case['queries']:
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input_lines.append(f"{q['type']} {q['l']} {q['r']}")
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problem = (
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"DZY给出了一个整数数组和m个查询，请处理以下两种操作：\n"
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"1. 类型1查询 (1 l r)：给区间[l, r]的每个元素a_i依次加上F_{i-l+1}（F为斐波那契数列，F₁=1, F₂=1）\n"
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"2. 类型2查询 (2 l r)：输出区间[l, r]元素和模1000000009\n\n"
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"输入格式：\n" +
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"\n".join(input_lines) +
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"\n\n请输出所有类型2查询的结果，每个结果占一行，并包裹在[answer]标签内，例如：\n"
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"[answer]\n17\n12\n[/answer]"
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)
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return problem
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@staticmethod
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def extract_output(output):
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import re
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answers = []
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# 匹配最后一个answer块内的所有数字
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matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
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if matches:
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last_match = matches[-1].strip()
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answers = [line.strip() for line in last_match.split('\n') if line.strip()]
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return answers if answers else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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# 将字符串答案转换为整数比较
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expected = identity['answers']
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return [int(ans) % MOD for ans in solution] == [x % MOD for x in expected]
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except (ValueError, KeyError, TypeError):
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return False
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