title: "CF 2025B - Binomial Coefficients, Kind Of"
description: "The code in the statement tries to build a Pascal triangle, but the recurrence is wrong. Instead of $$C[n][k] = C[n-1][k] + C[n-1][k-1]$$ it computes $$C[n][k] = C[n][k-1] + C[n-1][k-1].$$ The boundary values are still initialized as $$C[n][0] = 1,qquad C[n][n] = 1." date: "2026-06-09T03:20:23+07:00" tags: ["codeforces", "competitive-programming", "combinatorics", "dp", "math"] categories: ["algorithms"] codeforces_contest: 2025 codeforces_index: "B" codeforces_contest_name: "Educational Codeforces Round 170 (Rated for Div. 2)" rating: 1100 weight: 2025 solve_time_s: 437 verified: false draft: false --- CF 2025B - Binomial Coefficients, Kind Of Rating: 1100 Tags: combinatorics, dp, math Solve time: 7m 17s Verified: no ## Solution ## Problem Understanding The code in the statement tries to build a Pascal triangle, but the recurrence is wrong. Instead of$$C[n][k] = C[n-1][k] + C[n-1][k-1]$$
it computes
$$C[n][k] = C[n][k-1] + C[n-1][k-1].$$
The boundary values are still initialized as
$$C[n][0] = 1,\qquad C[n][n] = 1.$$
For many queries $(n,k)$, we must determine the value produced by this incorrect recurrence and print it modulo $10^9+7$.
The first input line gives the number of queries. The second line contains all $n_i$ values and the third line contains the corresponding $k_i$ values. For every pair $(n_i,k_i)$, we must output the entry that appears in the wrongly generated triangle.
The largest $n$ is $10^5$, and there can also be $10^5$ queries. Any algorithm that explicitly constructs a triangle of size $10^5$ is impossible. A triangle with $10^5$ rows contains roughly $5 \cdot 10^9$ cells, far beyond both the time and memory limits.
The constraints strongly suggest that there is a simple closed-form expression for the values. Once that expression is found, preprocessing up to $10^5$ and answering each query in $O(1)$ time becomes feasible.
A subtle trap is assuming that the wrong recurrence still produces ordinary binomial coefficients. For example:
Input:
1 5 2
The correct binomial coefficient would be
$$\binom{5}{2}=10,$$
but the wrong triangle gives 4, so directly using combinations produces the wrong answer.
Another easy mistake is to derive a formula involving powers of two but forget the modulo operation. For large $k$, values grow exponentially. For example, $k=99999$ requires modular exponentiation. Ordinary integer powers would be far too large.
A third trap is trying to build rows up to the queried $n$. The answer actually turns out to depend only on $k$, so any solution that processes all rows up to $10^5$ for every query will time out.
## Approaches
The most direct approach is to simulate exactly what the buggy code does.
We initialize $C[n][0]=1$ and $C[n][n]=1$, then fill every interior position using the given recurrence. This reproduces the intended triangle exactly, so it is obviously correct.
The problem is scale. To answer queries up to $n=10^5$, we would need to construct all rows up to $10^5$. The number of entries is
$$1+2+\cdots+100000 \approx 5 \times 10^9,$$
which is completely infeasible.
To get further, we need to understand the structure of the recurrence.
Let us write out the first few rows:
| n | Row values |
| --- | --- |
| 0 | 1 |
| 1 | 1 1 |
| 2 | 1 2 1 |
| 3 | 1 2 4 1 |
| 4 | 1 2 4 8 1 |
| 5 | 1 2 4 8 16 1 |
A pattern appears immediately. Every interior value seems to be a power of two.
Suppose every row above satisfies
$$C[n-1][k]=2^k$$
for interior positions.
Then
$$C[n][k]
=
C[n][k-1]+C[n-1][k-1].$$
If $C[n][k-1]=2^{k-1}$ and $C[n-1][k-1]=2^{k-1}$, then
$$C[n][k]
=
2^{k-1}+2^{k-1}
=
2^k.$$
This proves by induction that every interior entry equals $2^k$.
The row index $n$ disappears entirely. As long as $1 \le k < n$,
$$C[n][k] = 2^k.$$
The whole problem reduces to answering powers of two modulo $10^9+7$.
Since $k\le 99999$, we can precompute
$$pow2[i]=2^i \bmod (10^9+7)$$
for all needed exponents and answer every query in constant time.
| Approach | Time Complexity | Space Complexity | Verdict |
| --- | --- | --- | --- |
| Brute Force triangle construction | $O(N^2)$ | $O(N^2)$ | Too slow |
| Power-of-two observation + preprocessing | $O(K_{\max}+t)$ | $O(K_{\max})$ | Accepted |
## Algorithm Walkthrough
1. Read all query pairs.
2. Find the largest queried value of $k$. This determines how many powers of two we must precompute.
3. Build an array pow2 where pow2[i] = 2^i mod M.
4. Initialize pow2[0] = 1.
5. For each i from 1 to max_k, compute
$$pow2[i] = 2 \cdot pow2[i-1] \pmod M.$$
This gives every required power of two in linear time.
6. For each query $(n,k)$, output pow2[k].
The derived formula shows that every interior entry of the buggy triangle equals $2^k$, independent of $n$.
### Why it works
We prove that every interior entry satisfies
$$C[n][k]=2^k$$
for $1\le k<n$.
The base case is $k=1$:
$$C[n][1]
=
C[n][0]+C[n-1][0]
=
1+1
=
2
=
2^1.$$
Assume for some $k>1$ that all interior entries with column $k-1$ equal $2^{k-1}$. Then
$$C[n][k]
=
C[n][k-1]+C[n-1][k-1]
=
2^{k-1}+2^{k-1}
=
2^k.$$
Thus every interior column $k$ equals $2^k$. Since the formula depends only on $k$, every query answer is simply $2^k$ modulo $10^9+7$.
## Python Solution
python import sys input = sys.stdin.readline MOD = 1000000007 t = int(input()) n = list(map(int, input().split())) k = list(map(int, input().split())) max_k = max(k) pow2 = [0] * (max_k + 1) pow2[0] = 1 for i in range(1, max_k + 1): pow2[i] = (pow2[i - 1] * 2) % MOD print(*[pow2[x] for x in k])
The solution never uses the queried values of $n$. The proof shows that every valid interior position depends only on the column index $k$.
The preprocessing array stores all powers of two that may be needed. Since the maximum possible $k$ is less than $10^5$, this requires only about one hundred thousand entries.
An easy implementation mistake is to compute powers separately for each query using repeated multiplication. That would still pass, but preprocessing is simpler and avoids redundant work. Another common mistake is forgetting the modulo at every multiplication step, which causes numbers to become enormous.
## Worked Examples
### Sample 1
Input:
7 2 5 5 100000 100000 100000 100000 1 2 3 1 33333 66666 99999
Precomputed values:
| k | 2^k mod M |
| --- | --- |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
| 33333 | 326186014 |
| 66666 | 984426998 |
| 99999 | 303861760 |
Query processing:
| Query | k | Answer |
| --- | --- | --- |
| (2,1) | 1 | 2 |
| (5,2) | 2 | 4 |
| (5,3) | 3 | 8 |
| (100000,1) | 1 | 2 |
| (100000,33333) | 33333 | 326186014 |
| (100000,66666) | 66666 | 984426998 |
| (100000,99999) | 99999 | 303861760 |
This example demonstrates the central observation: even though the row indices vary dramatically, the answer depends only on $k$.
### Constructed Example
Input:
3 3 4 6 1 2 5
Preprocessing:
| k | 2^k |
| --- | --- |
| 1 | 2 |
| 2 | 4 |
| 5 | 32 |
Queries:
| Query | k | Answer |
| --- | --- | --- |
| (3,1) | 1 | 2 |
| (4,2) | 2 | 4 |
| (6,5) | 5 | 32 |
Output:
2 4 32
This trace shows that even positions very close to the right boundary still follow the same formula.
## Complexity Analysis
| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | $O(\max k + t)$ | Precompute powers once, then answer each query in constant time |
| Space | $O(\max k)$ | Store all powers of two up to the largest queried exponent |
With $\max k < 10^5$ and $t \le 10^5$, both time and memory usage are tiny compared to the limits.
## Test Cases
python # helper: run solution on input string, return output string import sys, io def run(inp: str) -> str: sys.stdin = io.StringIO(inp) MOD = 1000000007 t = int(input()) n = list(map(int, input().split())) k = list(map(int, input().split())) mx = max(k) pow2 = [0] * (mx + 1) pow2[0] = 1 for i in range(1, mx + 1): pow2[i] = (pow2[i - 1] * 2) % MOD return " ".join(str(pow2[x]) for x in k) # provided sample assert run( """7 2 5 5 100000 100000 100000 100000 1 2 3 1 33333 66666 99999 """ ) == "2 4 8 2 326186014 984426998 303861760" # minimum valid input assert run( """1 2 1 """ ) == "2" # several rows, same k assert run( """4 3 4 10 100 1 1 1 1 """ ) == "2 2 2 2" # small powers assert run( """5 2 3 4 5 6 1 2 3 4 5 """ ) == "2 4 8 16 32" # boundary near right edge assert run( """2 5 10 4 9 """ ) == "16 512"
| Test input | Expected output | What it validates |
| --- | --- | --- |
| n=2, k=1 | 2 | Smallest valid query |
| Many different rows, same k=1 | 2 2 2 2 | Independence from n |
| Consecutive small powers | 2 4 8 16 32 | Correct power generation |
| (5,4) and (10,9) | 16 512 | Values near the right boundary |
## Edge Cases
Consider the smallest possible query:
1 2 1
The recurrence gives
$$C[2][1]=C[2][0]+C[1][0]=1+1=2.$$
The algorithm returns $2^1=2$, which matches exactly.
Consider a query near the right boundary:
1 5 4
A manual construction gives row 5 as
$$[1,2,4,8,16,1].$$
The answer is 16. The algorithm returns $2^4=16$, so being close to the boundary does not change the formula.
Consider two queries with the same column but different rows:
2 10 100000 3 3
The recurrence produces 8 in column 3 regardless of the row. The algorithm outputs $2^3=8$ for both queries. This confirms that the row index plays no role once the formula has been derived.
Finally, consider the largest exponent:
1 100000 99999
The actual value is astronomically large. The algorithm never stores that integer directly. It computes powers modulo $10^9+7$ during preprocessing, producing the required result efficiently within the limits.