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To help improve her math, Kudryavka is given a set...

Created on: August 10, 2025

Answered using GPT-5 Thinking by Chat01

Question

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≤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 109+7
.

Input
Each test contains multiple test cases. The first line contains the number of test cases t
(1≤t≤104
). The description of the test cases follows.

The first line of each test case contains two integers n
and k
(1≤n≤2⋅105
, 1≤k≤109
).

The second line of each test case contains n
integers s1,s2,…,sn
(1≤si≤109
, si≠sj
) — 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⋅105
.

Output
For each test case, output an integer indicating the answer modulo 109+7
.

one thing I noticed removing x from S will lead to next 2^x-1 extra operations like this 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1 when x=5

can you code me in java

Python

python
# Let's simulate the process and compare with block-based product method.
from collections import Counter
MOD=10**9+7

def simulate(S, k):
    S=set(S)
    score=1
    for step in range(k):
        m=min(S)
        score = (score * m) % MOD
        S.remove(m)
        for i in range(1, m):
            if i not in S:
                S.add(i)
    return score

def block_product(S, k):
    S=sorted(S)
    rem=k
    ans=1
    # precompute B[m] for m<=29
    B=[1]*30
    for m in range(1,30):
        prod=1
        for r in range(1, m+1):
            prod = (prod * pow(r, 1<<(m-r), MOD)) % MOD
        B[m]=prod
    # process blocks
    for a in S:
        if rem==0: break
        m=a-1
        if m<0: 
            raise
        if m>=30:
            # partial always
            c=rem
            ans = (ans * a) % MOD  # first element
            X=c-1
            if X>0:
                maxr = min(m, X.bit_length())
                for r in range(1, maxr+1):
                    cnt = X//(1<<(r-1)) - X//(1<<r)
                    if cnt:
                        ans = (ans * pow(r, cnt, MOD)) % MOD
            rem=0
            break
        block_len = 1<<m
        if block_len<=rem:
            ans = (ans * a) % MOD
            ans = (ans * B[m]) % MOD
            rem -= block_len
        else:
            # partial
            c=rem
            ans = (ans * a) % MOD
            X=c-1
            if X>0:
                maxr = min(m, X.bit_length())
                for r in range(1, maxr+1):
                    cnt = X//(1<<(r-1)) - X//(1<<r)
                    if cnt:
                        ans = (ans * pow(r, cnt, MOD)) % MOD
            rem=0
            break
    return ans

# test random small cases
import random
def test():
    for _ in range(2000):
        n=random.randint(1,5)
        S=random.sample(range(1,12), n)  # small values to simulate
        # ensure k <= total operations, which equals sum over blocks length 2^{a-1}
        total=sum(1<<(a-1) for a in sorted(S))
        for k in range(1, min(40, total)+1):
            s=simulate(S, k)
            b=block_product(S, k)
            if s!=b:
                return S,k,s,b,total
    return "All good"
test()

'All good'

Thought

Algorithm breakdown

The algorithm initializes some constants and processes test cases by sorting the array S. It then attempts to consume elements of the array based on predefined rules and block lengths. If any remaining quantities can't fit within the block, partial computation will occur. Key edge cases are checked for ensuring the solution works efficiently and the constraints hold. Elements of the array are processed based on their magnitude, using modular arithmetic and exponentiation as needed.

Implementing in Java

The decomposition and formula seem correct, so now it's about translating it into Java. I'll first read the number of test cases and loop through them. For each test case, I’ll handle the array, sort it, and precompute some values globally. Then, I'll carefully implement the logic for each element, especially handling the larger values of rem and m.

Final adjustments

I’m thinking about the block length and how it scales with increasing m. For m ≥ 30, the block length will always be larger than k, so avoiding overflow is important. I can calculate the necessary bits using bitLength(X) and handle cnt calculations carefully, ensuring no integer overflows occur.

Handling edge cases

I’m considering edge cases where a might be 1, making m=0. In this case, block_len = 1, and the block product simplifies to a * B[0]. Since B[0] = 1 by definition, this is perfectly fine. For each condition, I'll apply the appropriate calculations to ensure correctness.

Thought for 3m 50s

Answer

you’re on the right track with the pattern—just one tiny correction:
after you remove x, the number of extra operations to exhaust the numbers < x is 2^(x-1) - 1 (not 2^x - 1).
e.g. for x = 5 the next pops are 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1 (length 2^4 - 1 = 15).

the key trick for fast math: sort the initial set a1 < a2 < … < an.
The pop sequence is block-by-block:

  • Block for ai has length 2^(ai-1) and equals: [ai] then a canonical sequence T(ai-1) of length 2^(ai-1)-1.
  • In any prefix of T(m), the value r (1 ≤ r ≤ m) appears
    floor(X/2^(r-1)) − floor(X/2^r) times in the first X terms.
    This means any partial block needs only O(log X) bases (≤ ~30 for X ≤ 1e9).

So we can eat whole blocks while k remains; once we enter a partial block we multiply ai and then the counts above for X = remaining - 1.

Here’s a clean, fast Java solution (O(n log k) per test). It never simulates the process:

java
import java.io.*;
import java.util.*;

public class Main {
    static final long MOD = 1_000_000_007L;
    static final int MAX_M_FULL = 29; // because 2^30 > 1e9 (k max)
    static long[] blockTailProd = new long[MAX_M_FULL + 1]; // B[m] = ∏_{r=1..m} r^(2^(m-r)) mod MOD

    public static void main(String[] args) throws Exception {
        FastScanner fs = new FastScanner(System.in);
        precomputeBlockTails();
        StringBuilder out = new StringBuilder();
        int t = fs.nextInt();
        while (t-- > 0) {
            int n = fs.nextInt();
            long k = fs.nextLong();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = fs.nextInt();
            Arrays.sort(a);

            long remaining = k;
            long ans = 1;

            for (int ai : a) {
                if (remaining == 0) break;
                int m = ai - 1; // T(m) length is 2^m - 1, block length is 2^m
                if (m >= MAX_M_FULL + 1) {
                    // This block is longer than any possible remaining (since k ≤ 1e9)
                    ans = mul(ans, ai);
                    long X = remaining - 1; // we will take X terms from T(m)
                    if (X > 0) ans = mul(ans, productPrefixT(m, X));
                    remaining = 0;
                    break;
                } else {
                    long blockLen = 1L << m; // safe for m ≤ 29
                    if (blockLen <= remaining) {
                        // take the full block: ai * B[m]
                        ans = mul(ans, ai);
                        ans = mul(ans, blockTailProd[m]);
                        remaining -= blockLen;
                    } else {
                        // take a partial prefix of length 'remaining'
                        ans = mul(ans, ai);
                        long X = remaining - 1; // terms taken from T(m)
                        if (X > 0) ans = mul(ans, productPrefixT(m, X));
                        remaining = 0;
                        break;
                    }
                }
            }
            // Problem guarantees enough operations exist (set is non-empty before each of the k ops)
            out.append(ans % MOD).append('\n');
        }
        System.out.print(out.toString());
    }

    // Precompute B[m] = ∏_{r=1..m} r^(2^(m-r)) for m = 0..29
    static void precomputeBlockTails() {
        blockTailProd[0] = 1; // T(0) empty
        for (int m = 1; m <= MAX_M_FULL; m++) {
            long prod = 1;
            for (int r = 1; r <= m; r++) {
                long exp = 1L << (m - r); // ≤ 2^28 for m ≤ 29
                prod = mul(prod, modPow(r, exp));
            }
            blockTailProd[m] = prod;
        }
    }

    // Product of first X terms of T(m): ∏_{r=1..min(m, floor(log2 X)+1)} r^{ floor(X/2^{r-1}) - floor(X/2^r) }
    static long productPrefixT(int m, long X) {
        long res = 1;
        long denomPrev = 1; // 2^{r-1}
        long denom = 2;     // 2^r
        for (int r = 1; r <= m && denomPrev <= X; r++) {
            long cnt = X / denomPrev - X / denom;
            if (cnt > 0) res = mul(res, modPow(r, cnt));
            denomPrev <<= 1;
            denom <<= 1;
        }
        return res;
    }

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

    static long modPow(long base, long exp) {
        long res = 1 % MOD;
        long p = base % MOD;
        long e = exp;
        while (e > 0) {
            if ((e & 1L) != 0) res = (res * p) % MOD;
            p = (p * p) % MOD;
            e >>= 1;
        }
        return res;
    }

    // Fast scanner
    static final class FastScanner {
        private final InputStream in;
        private final byte[] buffer = new byte[1 << 16];
        private int ptr = 0, len = 0;
        FastScanner(InputStream is) { in = is; }
        private int read() throws IOException {
            if (ptr >= len) {
                len = in.read(buffer);
                ptr = 0;
                if (len <= 0) return -1;
            }
            return buffer[ptr++];
        }
        long nextLong() throws IOException {
            long sgn = 1, x = 0;
            int c = read();
            while (c <= ' ') { if (c == -1) return Long.MIN_VALUE; c = read(); }
            if (c == '-') { sgn = -1; c = read(); }
            while (c > ' ') { x = x * 10 + c - '0'; c = read(); }
            return x * sgn;
        }
        int nextInt() throws IOException { return (int) nextLong(); }
    }
}

Why this works (quick recap)

  • The pop order splits into blocks by the original sorted elements.
    Block for ai has length 2^(ai-1) and equals [ai] + T(ai-1).
  • For any prefix of T(m), the count of a value r is ⌊X/2^(r-1)⌋ − ⌊X/2^r⌋ — i.e., how many numbers t ≤ X have exactly r-1 trailing zeros.
  • Since k ≤ 1e9, any partial block needs at most ~30 bases, so we can multiply them fast with modular exponentiation.

If you want, I can add a tiny self-check harness or comments to print intermediate counts for debugging.

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