从另外给定的数字形成所有可能的数字 [英] All possible numbers formed from addition of given digits

问题描述

如果我有 NR 的数字，从 1到n ，其中研究号之间，我又如何能计算，可以从另外这些数字来形成所有可能的号码（无论是在2/3/4/5/6组...）缺少研究。

If I have n-r numbers, from 1 to n where r numbers are missing in between, then how can I calculate all possible numbers that can be formed from addition of these numbers (either in groups of 2/3/4/5/6...).

举例来说，可以说我有 5-2 号， 也就是说， 1 2 4 和 3.5 失踪。现在，我可以形成

For example, lets say I have 5-2 numbers, that is, 1 2 4 and 3 5 are missing. Now, I can form

1 - {1}
2 - {2}
3 - {1,2}
4 - {4}
5 - {1,4}
6 - {4,2}
7 - {1,2,4}
8 - Cannot be formed

这是我需要找出，也就是从1，我不能形成使用给定的数字组合的第一个数字。一个简单的逻辑将做精。谢谢！

This is I need to find out, that is the first number from 1 which I cannot form using the combination of the given digits. A simple logic would do fine. Thanks!

推荐答案

对于输入设置 s或多集取值只包含正整数，叶夫根Kluev的解决方案是的 O（2N）的，[我的其他解决方案]是的 O（nlogn）的。

For input sets or multisets which only contain positive integers, Evgeny Kluev's Solution is O(2n), [my other solution] is O(nlogn).

对于一个给定的输入号■您可以找到unformable数如下：

For a given input numbers you can find the unformable number as follows:

1. 收集号到 int_log [I] ，其中 I = INT元素（LOG2 （元素））
2. 找到 S [] 这是在preFIX金额 int_log []
3. 第一unformable数为 S [X] + 1 ，其中 int_log [X + 1] 不包含元素小于或等于 S [X] + 1

1. Collect elements of number into int_log[i] where i = int(log2(element))
2. Find S[] which is the prefix sum of int_log[]
3. The first unformable number is S[x] + 1 where int_log[x + 1] contains no element less than or equal to S[x] + 1

有关证明，在范围内的所有号码[0， S [I] ] 的可以通过子集来形成 int_log 小于2 ^I-1 看到：<一href="http://math.stackexchange.com/questions/1099359/0-sum-re$p$psentable-by-numbers-in-a-set">http://math.stackexchange.com/questions/1099359/0-sum-re$p$psentable-by-numbers-in-a-set

For a proof that all numbers in the range [0, S[i]] can be formed by subset of int_log smaller than 2^i-1 see: http://math.stackexchange.com/questions/1099359/0-sum-representable-by-numbers-in-a-set

set<int> numbers{ 1, 2, 5, 6, 7 };
unordered_multimap<unsigned long, decltype(numbers)::key_type> int_log;

for (auto& value : numbers){
    auto key = decltype(int_log)::key_type{};

    _BitScanReverse(&key, value); //If using gcc you'll need to use __builtin_clz here
    int_log.insert(make_pair(key, value));
}

auto result = decltype(numbers)::key_type{}; // Because we only ever need to look at the previous prefix sum value we'll use previousSum instead of a S[] here
auto sum = decltype(numbers)::key_type{};
auto i = decltype(int_log)::key_type{};
auto smallest = decltype(numbers)::key_type{};

do{
    result = sum;
    smallest = 2 << i;

    for (auto range = int_log.equal_range(i); range.first != range.second; ++range.first){
        const auto current = range.first->second;

        sum += current;
        smallest = min(smallest, current);
    }
    ++i;
} while (result >= smallest - 1);
cout << "First number that cannot be formed: " << result + 1 << endl;

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