[UCSD白板题] Pairwise Distinct Summands

Problem Introduction

This is an example of a problem where a subproblem of the corresponding greedy algorithm is slightly distinct from the initial problem.

Problem Description

Task.The goal of this problem is to represent a given positive integer \(n\) as a sum of as many pairwise distinct positive integers as possible. That is, to find the maximum \(k\) such that \(n\) can be written as \(a_1+a_2+\cdots+a_k\) where \(a_1, \cdots, a_k\) are positive integers and \(a_i \neq a_j\) for all \(1 \leq i < j \leq k\).

Input Format.The input consists of a single integer \(n\).

Constraints.\(1 \leq n \leq 10^9\).

Output Format.In the first line, output the maximum number \(k\) such that \(n\) can be represented as a sum of \(k\) pairwise distinct positive integers. In the second line, output \(k\) pairwise distinct positive integers that sum up tp \(n\)(if there are many such representation, output any of them).

Sample 1.
Input:

6

Output:

3
1 2 3

Sample 2.
Input:

8

Output:

3
1 2 5

Sample 3.
Input:

2

Output:

1
2

算法分析

引理： 整数\(k\)由\(p\)个不重复的被加数组成，每一项至少为\(l\)，令\(k>2l\)并让这样的\(p\)取最大值。那么存在一个最佳的表示方式\(k=a_1+a_2+\cdots+a_p\)(每一项都不小于\(l\)并且两两不同)使得\(a_1=l\)。

证明：考虑某种最佳的表示方式\(k=b_1+b_2+\cdots+b_p\)。不失一般性，不妨假设\(b_1<b_2<\cdots<b_p\)，已知\(p\geq2\)（因为\(k>2l\)）。如果\(b_1=l\)，那么结论成立。否则，令\(\Delta=b_1-l \geq 1\)，考虑以下的表示方式：\(n=(b_1-\Delta)+b2+\cdots+(b_p+\Delta)\)，不难发现，这是一个最佳的表示方式（包括p个被加数并且两两不同）。

Solution

# Uses python3
import sys

def optimal_summands(n):
    summands = []
    k, l = n, 1
    while k > 2 * l:
        summands.append(l)
        k, l = k-l, l+1
    summands.append(k)
    return summands

if __name__ == '__main__':
    input = sys.stdin.read()
    n = int(input)
    summands = optimal_summands(n)
    print(len(summands))
    for x in summands:
        print(x, end=' ')