问倒置计数的除法和征服算法

EN

#include <iostream>
#include <tuple>
#include <vector>

using IntVec = std::vector<int>;

std::tuple<IntVec, long long> count_inv(IntVec xs);
std::tuple<IntVec, long long> count_split_inv(IntVec xs, IntVec ys);

std::tuple<IntVec, long long> count_inv(IntVec xs) {
    int n = xs.size();

    if (n < 2) return std::make_tuple(xs, 0);

    IntVec::iterator middle = xs.begin() + (n / 2);
    IntVec left(xs.begin(), middle);
    IntVec right(middle, xs.end());

    long long left_inv, right_inv, split_inv;
    IntVec as, bs, cs;
    std::tie(as, left_inv) = count_inv(left);
    std::tie(bs, right_inv) = count_inv(right);
    std::tie(cs, split_inv) = count_split_inv(as, bs);
    return std::make_tuple(cs, left_inv + right_inv + split_inv);
}

std::tuple<IntVec, long long> count_split_inv(IntVec xs, IntVec ys) {
    int n = xs.size() + ys.size();
    IntVec zs(n);
    long long split_inv = 0;

    int i = 0, j = 0, k = 0;
    for (; k < n && i < xs.size() && j < ys.size(); ++k) {
        if (xs[i] <= ys[j]) {
            zs[k] = xs[i];
            ++i;
        } else {
            zs[k] = ys[j];
            ++j;
            split_inv += xs.size() - i;
        }
    }

    for (; i < xs.size(); ++k, ++i)
        zs[k] = xs[i];

    for (; j < ys.size(); ++k, ++j)
        zs[k] = ys[j];

    return std::make_tuple(zs, split_inv);
}

int main() {
    std::vector<int> xs;
    int x;
    while (std::cin >> x) {
        xs.push_back(x);
    }

    IntVec ys;
    long long inv;
    std::tie(ys, inv) = count_inv(xs);
    std::cout << inv << std::endl;
    return 0;
}