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G. Wafu! time limit per test3 seconds memory limit...

Created on: August 10, 2025

Answered using GPT-5 Thinking by Chat01

Question

G. Wafu!
time limit per test3 seconds
memory limit per test256 megabytes

To help improve her math, Kudryavka is given a set S that consists of n distinct positive integers.

Initially, her score is 1. She can perform an arbitrary number of the following operations on the set if it is not empty:

Let the minimum value of S be m.
Multiply her score by m.
Remove m from S.
For every integer i such that 1 \le i < m, add i to the set S. It can be shown that no duplicates are added during this step.
She is addicted to performing operations, but after k operations, she realizes she forgot her score. Please help her determine her score, modulo 10^9+7.

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 \le t \le 10^4). The description of the test cases follows.

The first line of each test case contains two integers n and k (1 \le n \le 2 \cdot 10^5, 1 \le k \le 10^9).

The second line of each test case contains n integers s_1, s_2, \dots, s_n (1 \le s_i \le 10^9, s_i \neq s_j) — the elements of the initial set S. It is guaranteed that the set S is not empty before each of the k operations is performed.

It is guaranteed that the sum of n over all test cases does not exceed 2 \cdot 10^5.

Output
For each test case, output an integer indicating the answer modulo 10^9+7.

Example
InputCopy
4
2 3
1 3
3 6
5 1 4
2 100
2 100
5 15
1 2 3 4 5
OutputCopy
3
24
118143737
576
Note
Let us simulate the process in the first test case:

{1,3} \xrightarrow{\text{remove}\ 1} {3} \xrightarrow[\text{add}\ 1,2]{\text{remove}\ 3} {1,2} \xrightarrow{\text{remove}\ 1} {2}

The removed values are 1, 3 and 1 respectively, so her score is 1\times 3\times 1 = 3.

In the second test case, the answer is 1 \times 4 \times 1 \times 2 \times 1 \times 3 = 24.

give solution in cpp. There should be no comments in the code

Thought

Evaluating "G. Wafu!" problem

I believe the problem is related to Codeforces Round #1666, titled "G. Wafu!" The task involves computing the product of removed minimums after k operations performed on a set. Each step removes the minimal element and modifies the set, adding integers less than it. The set evolves into a consecutive prefix. Let's think through the sequence of operations carefully!

Exploring the "G. Wafu!" operations

After removing the minimum m, we add all integers from 1 to m-1, but only those not already in the set. This process results in a new set where the minimum becomes 1 (by default, unless 1's already in the set). After removing 1, we add nothing, and the cycle repeats but with a new minimum starting from 2.

Simulating the set's behavior

The minimum numbers in the set seem to bounce between 1 and increasing integers until the set doesn’t contain anything below a threshold M. The process eventually includes all integers from 1 to T. We’re processing k operations, and after each removal, integers less than x reappear in the set, with 1 reappearing frequently.

Understanding the cycle of removals

The sequence of removals keeps bouncing between 1 and the next minimum greater than 1. Initially, the set contains {1..t} and larger elements. After removing 1, the set becomes {2..t} and then the next minimum is 2. Removing 2 adds 1 back, which restarts the cycle until 3 becomes the minimum again.

Simulating the removal process

I’m trying to deduce the behavior of the set. When we start with {M}, the operations seem to follow a repeating cycle after a few steps. For example, with M=3, we perform 4 steps before the set empties. For M=4, this extends to 8 operations. It seems that the number of steps before emptiness could be 2^(M-1).

Testing sequence patterns

I’m exploring how the removal process works for various sets, starting with step counts for M=3, M=4, and others. After calculating the removal sequence, I’m working through different scenarios where I predict counts for each element. Ultimately, it seems the process follows a recursive structure when elements are added, and sorting the set seems to be key.

Analyzing removal patterns

I’m working through how each element in a set is removed in a structured sequence. It’s like a recursive process where each large value triggers a block of operations that mirrors the smaller values’ removal. Initially, we remove the smallest and then proceed to the next largest element, building patterns step by step until the set is empty.

Optimizing product length calculation

I’m trying to calculate a product for a sequence with recursive structures. As large numbers lead to huge sequence lengths, I’m exploring how to efficiently compute products for large values, focusing on the first k removals. The product length is determined by the recursive breakdown, and speeding this up requires binary-decomposition techniques to avoid deep recursion.

Optimizing recursion depth

I'm looking at reducing the recursion depth for large values of t. By using logarithmic techniques, I can quickly determine a cap for t, making it manageable, even for extremely large values of t. This helps avoid deep recursion by limiting t to a maximum of 31, streamlining the calculation.

Implementing prefix product

The algorithm involves calculating the prefix product using recursion. First, we check cases where L <= 0 or t == 0. For larger values, we compute the prefix product for smaller values of t, leveraging precomputed values for efficiency. We also ensure boundary cases like label values are handled safely, keeping the process better optimized while respecting size constraints.

Implementing prefix_prod function

I’m setting up a function that will handle prefix products efficiently by setting a cap for t based on the logarithmic values of L. I have to account for edge cases like when L == 0 or when ai == 1 (since P(0) is empty). The complexity seems manageable, but I’ll make sure the recursion and product arrays are correctly computed.

Validating algorithm and constraints

I’m testing the algorithm for various cases and comparing against expected outputs. The logic seems to hold well in situations where n and k are manipulated through sorting and the prefix product is computed accordingly. I also want to ensure that performance is optimized for large values, particularly for L values close to 1e9. Precomputing arrays up to t=60 seems sensible for safety.

Ensuring recursive logic and edge cases

I’m finalizing the logic to ensure that the recursion handles edge cases properly, especially when L exceeds sizeN[t]. I should be mindful that after capping t, recursion computes results at most O(log L) times, and will handle large values safely. I also need to implement it all in C++ with modular multiplication adjustments, checking for overflow issues.

Thought for 3m 50s

Answer

Here’s a clean C++ solution that handles huge kk efficiently by exploiting the recursive structure of the sequence produced by the operation:

cpp
#include <bits/stdc++.h>
using namespace std;

const long long MOD = 1000000007LL;
const int TMAX = 60;
const long long LIM = (long long)4e18;

long long sz[TMAX + 1];
long long prodAll[TMAX + 1];

long long mul(long long a, long long b){ return (a % MOD) * (b % MOD) % MOD; }

long long prefProd(int t, long long L){
    if (L <= 0 || t <= 0) return 1;
    if (t > TMAX){
        int lg = 63 - __builtin_clzll(L + 1);
        t = min(TMAX, lg + 1);
    }
    if (L >= sz[t]) return prodAll[t];
    long long left = sz[t - 1];
    if (L <= left) return prefProd(t - 1, L);
    if (L == left + 1) return mul(prodAll[t - 1], t);
    long long res = mul(prodAll[t - 1], t);
    return mul(res, prefProd(t - 1, L - left - 1));
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    sz[0] = 0;
    prodAll[0] = 1;
    for (int i = 1; i <= TMAX; ++i){
        sz[i] = sz[i - 1] + sz[i - 1] + 1;
        if (sz[i] > LIM) sz[i] = LIM;
        prodAll[i] = mul(mul(prodAll[i - 1], prodAll[i - 1]), i);
    }
    
    int T;
    if(!(cin >> T)) return 0;
    while (T--){
        int n;
        long long k;
        cin >> n >> k;
        vector<long long> a(n);
        for (int i = 0; i < n; ++i) cin >> a[i];
        sort(a.begin(), a.end());
        
        long long ans = 1;
        for (int i = 0; i < n && k > 0; ++i){
            ans = mul(ans, a[i] % MOD);
            --k;
            if (k == 0) break;
            long long t = a[i] - 1;
            if (t <= 0) continue;
            long long maxBlock = (t > TMAX ? LIM : sz[(int)t]);
            long long L = min(k, maxBlock);
            if (L > 0){
                ans = mul(ans, prefProd((int)min<long long>(t, TMAX), L));
                k -= L;
            }
        }
        cout << ans % MOD << "\n";
    }
    return 0;
}

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