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：
1 #include <iostream> 2 #include <set> 3 #include <vector> 4 using namespace std; 5 6 typedef long long ll; 7 8 inline bool check(ll x, ll a, ll b) 9 { 10 return x * (x + 1) / 2 <= a + b; 11 } 12 13 int main() 14 { 15 ll a, b; 16 while (cin >> a >> b) 17 { 18 ll l = 1, r = 200000, ans = -1; 19 while (l <= r) 20 { 21 ll m = l + r >> 1; 22 if (check(m, a, b)) 23 { 24 l = m + 1; 25 ans = m; 26 } 27 else 28 { 29 r = m - 1; 30 } 31 } 32 set<int> st; 33 int x = min(a, b); 34 for (int i = ans; i >= 1; i--) 35 { 36 if (i > x) continue; 37 st.insert(i); 38 x -= i; 39 } 40 set<int> st2; 41 for (int i = 1; i <= ans; i++) 42 { 43 if (!st.count(i)) st2.insert(i); 44 } 45 if (a > b) swap(st, st2); 46 cout << st.size() << endl; 47 for (auto it: st) cout << it << " "; 48 cout << endl; 49 cout << st2.size() << endl; 50 for (auto it: st2) cout << it << " "; 51 cout << endl; 52 } 53 return 0; 54 }