$A36. 优秀的拆分\ \ \ \ \ \begin{aligned} \colorbox{#ffdbdd}{\color{#ff6aa4}\texttt{ 入门 }} \ \end{aligned}\begin{aligned} \colorbox{#eaf2ff}{\color{#007bff}\texttt{ CSP-J }} \end{aligned}$

自制讲解视频

注：思路和解析都在视频中

---

20分 Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;

vector<int> a;
bool dfs(int cnt, int n, int ln) {
    if (cnt == n + 1) {
        ll sum = 0;
        for (auto &i : a)
            sum += pow(2, i);
        if (sum == n)
            for (int i = a.size() - 1; i >= 0; i ++)
                cout << pow(2, a[i]) << " ";
        return sum == n;
    }
    for (int i = 1; i <= ln; i ++) {
        a.push_back(i);
        if (dfs(cnt + 1, n, ln)) return 1;
        a.pop_back();
    }
    
    return 0;
}
int main() {
    int n;
    cin >> n;
    bool h = dfs(1, n, log2(n));
    if (!h) cout << -1;

    return 0;
}

---

60分 Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;

vector<int> a;
void dfs(int cnt, int n, int ln) {
    if (cnt == n + 1) {
        ll sum = 0;
        for (auto &i : a)
            sum += pow(2, i);
        if (sum == n)
            for (int i = a.size() - 1; i >= 0; i ++)
                cout << pow(2, a[i]) << " ";
    }
    for (int i = 1; i <= ln; i ++) {
        a.push_back(i);
        dfs(cnt + 1, n, ln);
        a.pop_back();
    }
    return;
}
int main() {
    int n;
    cin >> n;
    if (n % 2) {
        cout << -1;
        return 0;
    }
    dfs(1, n, log2(n));
    

    return 0;
}

---

AC Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
    int n;
    cin >> n;
    if (n % 2) {
        cout << -1;
        return 0;
    }
    int cnt = 0;
    vector<int> a;
    while (n) {
        if (n % 2)
            a.push_back(pow(2, cnt));
        n /= 2;
        cnt ++;
    }
    for (int i = a.size() - 1; i >= 0; i --)
        cout << a[i] << " ";

    return 0;
}

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