CF1072C Cram Time

思路：

首先二分找到使x * (x + 1) / 2 <= a + b最大的x，然后令p = min(a, b), q = max(a, b),按照x，x - 1, ..., 1的顺序选取数字把p填满，把剩下未选择的数字放到q中即可。

实现：

#include <iostream>
#include <set>
#include <vector>
using namespace std;

typedef long long ll;

inline bool check(ll x, ll a, ll b)
{
    return x * (x + 1) / 2 <= a + b;
}

int main()
{
    ll a, b;
    while (cin >> a >> b)
    {
        ll l = 1, r = 200000, ans = -1;
        while (l <= r)
        {
            ll m = l + r >> 1;
            if (check(m, a, b))
            {
                l = m + 1;
                ans = m;
            }
            else
            {
                r = m - 1;
            }
        }
        set<int> st;
        int x = min(a, b);
        for (int i = ans; i >= 1; i--)
        {
            if (i > x) continue;
            st.insert(i);
            x -= i;
        }
        set<int> st2;
        for (int i = 1; i <= ans; i++)
        {
            if (!st.count(i)) st2.insert(i);
        }
        if (a > b) swap(st, st2);
        cout << st.size() << endl;
        for (auto it: st) cout << it << " ";
        cout << endl;
        cout << st2.size() << endl;
        for (auto it: st2) cout << it << " ";
        cout << endl;
    }
    return 0;
}