mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-19 12:58:04 +00:00
331 lines
11 KiB
Python
Executable file
331 lines
11 KiB
Python
Executable file
"""#
|
||
|
||
### 谜题描述
|
||
Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types:
|
||
|
||
1. 1 l r x — increase all integers on the segment from l to r by values x;
|
||
2. 2 l r — find <image>, where f(x) is the x-th Fibonacci number. As this number may be large, you only have to find it modulo 109 + 7.
|
||
|
||
|
||
|
||
In this problem we define Fibonacci numbers as follows: f(1) = 1, f(2) = 1, f(x) = f(x - 1) + f(x - 2) for all x > 2.
|
||
|
||
Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha?
|
||
|
||
Input
|
||
|
||
The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of elements in the array and the number of queries respectively.
|
||
|
||
The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).
|
||
|
||
Then follow m lines with queries descriptions. Each of them contains integers tpi, li, ri and may be xi (1 ≤ tpi ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 109). Here tpi = 1 corresponds to the queries of the first type and tpi corresponds to the queries of the second type.
|
||
|
||
It's guaranteed that the input will contains at least one query of the second type.
|
||
|
||
Output
|
||
|
||
For each query of the second type print the answer modulo 109 + 7.
|
||
|
||
Examples
|
||
|
||
Input
|
||
|
||
5 4
|
||
1 1 2 1 1
|
||
2 1 5
|
||
1 2 4 2
|
||
2 2 4
|
||
2 1 5
|
||
|
||
|
||
Output
|
||
|
||
5
|
||
7
|
||
9
|
||
|
||
Note
|
||
|
||
Initially, array a is equal to 1, 1, 2, 1, 1.
|
||
|
||
The answer for the first query of the second type is f(1) + f(1) + f(2) + f(1) + f(1) = 1 + 1 + 1 + 1 + 1 = 5.
|
||
|
||
After the query 1 2 4 2 array a is equal to 1, 3, 4, 3, 1.
|
||
|
||
The answer for the second query of the second type is f(3) + f(4) + f(3) = 2 + 3 + 2 = 7.
|
||
|
||
The answer for the third query of the second type is f(1) + f(3) + f(4) + f(3) + f(1) = 1 + 2 + 3 + 2 + 1 = 9.
|
||
|
||
Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
|
||
```python
|
||
#include <bits/stdc++.h>
|
||
using namespace std;
|
||
inline int readInt() {
|
||
static int n, ch;
|
||
n = 0, ch = getchar();
|
||
while (!isdigit(ch)) ch = getchar();
|
||
while (isdigit(ch)) n = n * 10 + ch - '0', ch = getchar();
|
||
return n;
|
||
}
|
||
const int MOD = 1000000007;
|
||
int t[2][2];
|
||
struct Matrix {
|
||
int a[2][2];
|
||
Matrix() { memset(a, 0, sizeof a); }
|
||
int *operator[](int i) { return *(a + i); }
|
||
const int *operator[](int i) const { return *(a + i); }
|
||
inline void operator*=(const Matrix &m1) {
|
||
memset(t, 0, sizeof t);
|
||
for (int i = 0; i < 2; ++i)
|
||
for (int k = 0; k < 2; ++k)
|
||
for (int j = 0; j < 2; ++j)
|
||
(t[i][j] += (long long)a[i][k] * m1.a[k][j] % MOD) %= MOD;
|
||
memcpy(a, t, sizeof t);
|
||
}
|
||
} I;
|
||
const int MAX_S = 64;
|
||
Matrix mem[MAX_S];
|
||
inline Matrix f(long long n) {
|
||
Matrix A;
|
||
A[0][0] = A[1][1] = 1;
|
||
for (int i = 0; i < MAX_S; ++i)
|
||
if ((n >> i) & 1) A *= mem[i];
|
||
return A;
|
||
}
|
||
void init() {
|
||
mem[0][0][0] = mem[0][0][1] = mem[0][1][0] = 1;
|
||
for (int i = 1; i < MAX_S; ++i) {
|
||
mem[i] = mem[i - 1];
|
||
mem[i] *= mem[i - 1];
|
||
}
|
||
}
|
||
const int MAX_N = 100000 + 3;
|
||
int n, m, a[MAX_N];
|
||
struct SegmentTree {
|
||
static const int MAX_NODE = (1 << 17) * 2 + 1;
|
||
struct Node {
|
||
Matrix matrix, tagMul;
|
||
} nodes[MAX_NODE];
|
||
inline void multiply(int o, Matrix c) {
|
||
Node &v = nodes[o];
|
||
v.tagMul *= c;
|
||
v.matrix *= c;
|
||
}
|
||
inline void pushDown(int o) {
|
||
multiply(((o)*2 + 1), nodes[o].tagMul);
|
||
multiply(((o)*2 + 2), nodes[o].tagMul);
|
||
nodes[o].tagMul = I;
|
||
}
|
||
inline Matrix merge(const Matrix &a, const Matrix &b) {
|
||
Matrix c;
|
||
c[0][0] = (a[0][0] + b[0][0]) % MOD;
|
||
c[0][1] = (a[0][1] + b[0][1]) % MOD;
|
||
c[1][0] = (a[1][0] + b[1][0]) % MOD;
|
||
c[1][1] = (a[1][1] + b[1][1]) % MOD;
|
||
return c;
|
||
}
|
||
void build(int o, int l, int r, int a[]) {
|
||
Node &v = nodes[o];
|
||
v.tagMul = I;
|
||
if (r - l == 1)
|
||
v.matrix = f(a[l]);
|
||
else {
|
||
build(((o)*2 + 1), l, (((l) + (r)) >> 1), a),
|
||
build(((o)*2 + 2), (((l) + (r)) >> 1), r, a);
|
||
v.matrix = merge(nodes[((o)*2 + 1)].matrix, nodes[((o)*2 + 2)].matrix);
|
||
}
|
||
}
|
||
void init(int n, int a[]) { build(0, 0, n, a); }
|
||
Matrix query(int o, int l, int r, int a, int b) {
|
||
if (l >= a && r <= b)
|
||
return nodes[o].matrix;
|
||
else {
|
||
pushDown(o);
|
||
Matrix res;
|
||
if ((((l) + (r)) >> 1) > a)
|
||
res = merge(res, query(((o)*2 + 1), l, (((l) + (r)) >> 1), a, b));
|
||
if ((((l) + (r)) >> 1) < b)
|
||
res = merge(res, query(((o)*2 + 2), (((l) + (r)) >> 1), r, a, b));
|
||
return res;
|
||
}
|
||
}
|
||
void modify(int o, int l, int r, int a, int b, Matrix x) {
|
||
if (r <= a || l >= b) return;
|
||
if (l >= a && r <= b)
|
||
multiply(o, x);
|
||
else {
|
||
pushDown(o);
|
||
if ((((l) + (r)) >> 1) > a)
|
||
modify(((o)*2 + 1), l, (((l) + (r)) >> 1), a, b, x);
|
||
if ((((l) + (r)) >> 1) < b)
|
||
modify(((o)*2 + 2), (((l) + (r)) >> 1), r, a, b, x);
|
||
nodes[o].matrix =
|
||
merge(nodes[((o)*2 + 1)].matrix, nodes[((o)*2 + 2)].matrix);
|
||
}
|
||
}
|
||
inline int query(int l, int r) { return query(0, 0, n, l, r)[1][0]; }
|
||
inline void modify(int l, int r, int c) { modify(0, 0, n, l, r, f(c)); }
|
||
} segmentTree;
|
||
int main() {
|
||
init();
|
||
n = readInt(), m = readInt();
|
||
for (int i = 0; i < n; ++i) a[i] = readInt();
|
||
I[0][0] = 1, I[1][1] = 1;
|
||
segmentTree.init(n, a);
|
||
while (m--) {
|
||
int type = readInt(), l = readInt() - 1, r = readInt();
|
||
if (type == 1)
|
||
segmentTree.modify(l, r, readInt());
|
||
else if (type == 2)
|
||
printf(\"%d\n\", segmentTree.query(l, r));
|
||
else
|
||
assert(false);
|
||
}
|
||
return 0;
|
||
}
|
||
```
|
||
|
||
|
||
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。
|
||
"""
|
||
|
||
from bootcamp import Basebootcamp
|
||
from bootcamp import Basebootcamp
|
||
import random
|
||
import re
|
||
|
||
class Esashaandarraybootcamp(Basebootcamp):
|
||
def __init__(self, **params):
|
||
self.n = params.get('n', 5)
|
||
self.m = params.get('m', 5)
|
||
self.max_initial = params.get('max_initial', 5)
|
||
self.max_x = params.get('max_x', 5)
|
||
|
||
def case_generator(self):
|
||
MOD = 10**9 +7
|
||
|
||
# Generate initial array
|
||
a_initial = [random.randint(1, self.max_initial) for _ in range(self.n)]
|
||
queries = []
|
||
diff = [0] * (self.n + 2) # 差分数组(1-based)
|
||
|
||
# Generate queries ensuring at least one type2
|
||
type2_count = 0
|
||
for _ in range(self.m):
|
||
# Force last query to be type2 if no type2 generated
|
||
if type2_count == 0 and len(queries) == self.m -1:
|
||
t = 2
|
||
else:
|
||
t = random.choice([1,2])
|
||
|
||
l = random.randint(1, self.n)
|
||
r = random.randint(l, self.n)
|
||
|
||
if t ==1:
|
||
x = random.randint(1, self.max_x)
|
||
queries.append({'type':1, 'l':l, 'r':r, 'x':x})
|
||
# 关键修复点：移除原条件判断
|
||
diff[l-1] += x
|
||
diff[r] -= x # 直接修改r位置，无需条件判断
|
||
else:
|
||
queries.append({'type':2, 'l':l, 'r':r})
|
||
type2_count +=1
|
||
|
||
# 计算delta数组的Prefix Sum
|
||
delta = [0]*self.n
|
||
current_diff =0
|
||
for i in range(self.n):
|
||
current_diff += diff[i]
|
||
delta[i] = current_diff
|
||
|
||
# 计算expected_outputs
|
||
expected_outputs = []
|
||
for q in queries:
|
||
if q['type'] ==2:
|
||
l, r = q['l'], q['r']
|
||
total =0
|
||
for i in range(l-1, r):
|
||
a = a_initial[i] + delta[i]
|
||
total = (total + self.fib_mod(a)) % MOD
|
||
expected_outputs.append(total)
|
||
|
||
# 确保至少一个type2
|
||
if not expected_outputs:
|
||
l, r = 1, self.n
|
||
total = sum(self.fib_mod(a_initial[i] + delta[i]) for i in range(self.n)) % MOD
|
||
expected_outputs.append(total)
|
||
queries[-1] = {'type':2, 'l':l, 'r':r}
|
||
|
||
return {
|
||
'n': self.n,
|
||
'm': self.m,
|
||
'initial_array': a_initial,
|
||
'queries': queries,
|
||
'expected_outputs': expected_outputs
|
||
}
|
||
|
||
@staticmethod
|
||
def fib_mod(x):
|
||
MOD = 10**9 +7
|
||
if x ==0:
|
||
return 0
|
||
if x ==1 or x ==2:
|
||
return 1 % MOD
|
||
|
||
def multiply(mat_a, mat_b):
|
||
return [
|
||
[(mat_a[0][0]*mat_b[0][0] + mat_a[0][1]*mat_b[1][0]) % MOD,
|
||
(mat_a[0][0]*mat_b[0][1] + mat_a[0][1]*mat_b[1][1]) % MOD],
|
||
[(mat_a[1][0]*mat_b[0][0] + mat_a[1][1]*mat_b[1][0]) % MOD,
|
||
(mat_a[1][0]*mat_b[0][1] + mat_a[1][1]*mat_b[1][1]) % MOD]
|
||
]
|
||
|
||
def matrix_power(mat, power):
|
||
res = [[1,0],[0,1]]
|
||
while power >0:
|
||
if power %2 ==1:
|
||
res = multiply(res, mat)
|
||
mat = multiply(mat, mat)
|
||
power //=2
|
||
return res
|
||
|
||
trans = [[1,1],[1,0]]
|
||
powered = matrix_power(trans, x-2)
|
||
return (powered[0][0] + powered[0][1]) % MOD
|
||
|
||
@staticmethod
|
||
def prompt_func(question_case) -> str:
|
||
problem_desc = "Sasha has an array of integers and needs to process several queries. Each query is of one of the following types:\n"
|
||
problem_desc += "1. Type 1: Increase all elements from position l to r by x.\n"
|
||
problem_desc += "2. Type 2: Compute the sum of Fibonacci numbers for elements from position l to r, modulo 1e9+7.\n"
|
||
problem_desc += "The Fibonacci numbers are defined as f(1)=1, f(2)=1, f(x)=f(x-1)+f(x-2) for x>2.\n\n"
|
||
problem_desc += f"Initial array: {question_case['initial_array']}\n\n"
|
||
problem_desc += f"Queries (total {len(question_case['queries'])}):\n"
|
||
for idx, q in enumerate(question_case['queries'], 1):
|
||
if q['type'] ==1:
|
||
problem_desc += f"{idx}. Type 1: l={q['l']}, r={q['r']}, x={q['x']}\n"
|
||
else:
|
||
problem_desc += f"{idx}. Type 2: l={q['l']}, r={q['r']}\n"
|
||
problem_desc += "\nFor each Type 2 query, compute the sum of Fibonacci numbers modulo 1e9+7. Provide all answers in the order of the queries, each on a new line enclosed within [answer] and [/answer]. Example:\n[answer]\n5\n7\n9\n[/answer]"
|
||
return problem_desc
|
||
|
||
@staticmethod
|
||
def extract_output(output):
|
||
matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
|
||
if not matches:
|
||
return None
|
||
last_match = matches[-1].strip()
|
||
solution = []
|
||
for line in last_match.split('\n'):
|
||
stripped = line.strip()
|
||
if stripped:
|
||
try:
|
||
solution.append(int(stripped))
|
||
except:
|
||
pass
|
||
return solution if solution else None
|
||
|
||
@classmethod
|
||
def _verify_correction(cls, solution, identity):
|
||
expected = identity.get('expected_outputs', [])
|
||
return solution == expected
|