CF 105669I - Partition
We are given a single integer ( n ). The task is to count how many different ways we can express ( n ) as a sum of positive integers, where order matters.
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Solve time: 37s
Verified: yes
Solution
Problem Understanding
We are given a single integer ( n ). The task is to count how many different ways we can express ( n ) as a sum of positive integers, where order matters. For example, writing ( 3 ) as ( 1+2 ) is considered different from ( 2+1 ), and even the single term ( 3 ) itself is also a valid representation. Every sequence of positive integers whose sum is exactly ( n ) contributes one way.
This is the classic problem of counting ordered integer compositions of ( n ). The answer grows quickly, and the final result must be reported modulo ( 10^9+7 ).
The constraint ( n \le 10^6 ) is the key signal here. Any approach that enumerates partitions or builds all compositions explicitly is impossible because even for moderate ( n ), the number of compositions is exponential in ( n ). The structure of the problem must therefore be reduced to a simple recurrence or closed-form identity computable in linear time.
A subtle edge case appears at ( n = 1 ). There is exactly one composition: ([1]). A naive interpretation that disallows using the number itself would incorrectly output zero. Another common mistake is to confuse ordered compositions with unordered partitions; for instance, for ( n=3 ), the correct answer is (4), not (3), because order is counted.
Approaches
A brute-force solution would try to generate every possible sequence of positive integers summing to ( n ). From a recursion standpoint, at each step we choose the next summand ( k \ge 1 ) and recurse on ( n-k ). This creates a branching factor of up to ( n ) at the top level, then ( n-1 ), and so on. The total number of recursive states expands roughly like the number of compositions itself, which is ( 2^{n-1} ). For ( n = 10^6 ), this is completely infeasible.
The key observation is that the problem does not depend on the actual values chosen, only on the remaining sum. Let ( f(n) ) be the number of valid compositions of ( n ). Any composition of ( n ) either starts with ( 1 ), or ( 2 ), or ( 3 ), and so on up to ( n ). This gives the recurrence [ f(n) = f(n-1) + f(n-2) + \dots + f(0), ] with the convention ( f(0) = 1 ), representing the empty decomposition.
A direct computation of this recurrence is still ( O(n^2) ) if implemented naively, but it can be simplified by noticing a telescoping relationship. Subtracting the recurrence for ( f(n-1) ) from that of ( f(n) ) yields [ f(n) - f(n-1) = f(n-1), ] which simplifies to [ f(n) = 2 \cdot f(n-1). ]
This collapses the entire structure into a simple doubling process: each integer composition of ( n-1 ) can either remain unchanged with a leading ( 1 ), or be extended in a way that preserves bijection with compositions of ( n ). This establishes a one-to-one correspondence that doubles the count at every step.
We therefore get the closed form: [ f(n) = 2^{n-1}. ]
This can be computed in linear time via a simple recurrence or in logarithmic time via modular exponentiation.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Enumeration | ( O(2^n) ) | ( O(n) ) | Too slow |
| Prefix-sum DP over states | ( O(n^2) ) | ( O(n) ) | Too slow |
| Closed-form ( 2^{n-1} ) | ( O(\log n) ) | ( O(1) ) | Accepted |
Algorithm Walkthrough
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We recognize that every valid construction of ( n ) corresponds to a sequence of cuts between integers ( 1 ) through ( n ). Between each adjacent pair of positions, we either place a cut or we do not, which determines whether numbers are grouped or separated. This converts the problem into a binary decision process over ( n-1 ) gaps.
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Since each of the ( n-1 ) positions independently has two choices, the total number of configurations is ( 2^{n-1} ). This is the central combinatorial transformation that avoids recursion entirely.
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We compute ( 2^{n-1} \bmod (10^9+7) ) using fast exponentiation. The exponent is large, so we repeatedly square the base and reduce the exponent by half at each step.
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The final value is returned as the answer.
Why it works
The invariant is that every composition of ( n ) corresponds uniquely to a subset of the ( n-1 ) possible separators between consecutive integers. Choosing a subset determines exactly where a segment ends, and thus uniquely determines a valid ordered partition. No two different subsets produce the same composition, and every composition corresponds to exactly one subset. This bijection guarantees correctness of the ( 2^{n-1} ) count.
Python Solution
import sys
input = sys.stdin.readline
MOD = 10**9 + 7
n = int(input().strip())
# number of compositions of n is 2^(n-1)
if n == 0:
print(1)
else:
print(pow(2, n - 1, MOD))
The implementation directly applies modular exponentiation. The special case ( n = 0 ) is included for completeness, though the problem guarantees ( n \ge 1 ).
The key simplification is avoiding any DP table or recursion. The entire structure of the problem reduces to counting independent binary choices across the gaps between integers.
Worked Examples
Example 1: n = 3
We track how compositions arise from the binary gap interpretation.
| Gap decisions (between 1 and 2, 2 and 3) | Resulting composition |
|---|---|
| cut, cut | 1 + 1 + 1 |
| cut, no cut | 1 + 2 |
| no cut, cut | 2 + 1 |
| no cut, no cut | 3 |
This produces 4 valid compositions, matching ( 2^{2} = 4 ).
This trace confirms that every subset of gaps corresponds to exactly one valid decomposition.
Example 2: n = 4
| Gap decisions | Composition |
|---|---|
| 1,1,1 | 1+1+1+1 |
| 1,1,0 | 1+1+2 |
| 1,0,1 | 1+2+1 |
| 1,0,0 | 1+3 |
| 0,1,1 | 2+1+1 |
| 0,1,0 | 2+2 |
| 0,0,1 | 3+1 |
| 0,0,0 | 4 |
There are 8 outcomes, matching ( 2^{3} = 8 ). The structure of independent gap decisions is fully exposed here.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | ( O(\log n) ) | fast exponentiation computes ( 2^{n-1} ) in logarithmic steps |
| Space | ( O(1) ) | only a few integers are maintained |
The constraint ( n \le 10^6 ) is easily handled because exponentiation under modulus is extremely fast even for large exponents. The solution runs well within both time and memory limits.
Test Cases
import sys, io
MOD = 10**9 + 7
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
n = int(sys.stdin.readline().strip())
if n == 0:
return "1"
return str(pow(2, n - 1, MOD))
# provided samples
assert run("1") == "1"
assert run("3") == "4"
# custom cases
assert run("2") == "2", "1+1, 2"
assert run("4") == "8", "check small correctness"
assert run("10") == str(pow(2, 9, MOD)), "power correctness"
assert run("1000000") == str(pow(2, 999999, MOD)), "large stress case"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 | 1 | minimal boundary |
| 2 | 2 | smallest non-trivial composition |
| 4 | 8 | exponential growth consistency |
| 1000000 | 2^999999 mod | performance and large exponent |
Edge Cases
n = 1
For input 1, the algorithm computes ( 2^{0} = 1 ). The only composition is the single element [1], so the result is correct. The gap interpretation has zero gaps, meaning exactly one configuration exists.
Very large n (up to 10^6)
For n = 1000000, the exponentiation step runs in ( O(\log n) ). The computation repeatedly squares the base 2 and reduces the exponent, never storing large intermediate values beyond the modulus. This ensures the algorithm remains stable and fast even at maximum input size.