輸入

2 1

輸出

1 2

說明

The sum of subintervals [1],[1,2][1],[1,2][1],[1,2] both satisfy ≡1(mod2), and their lengths are1,21,21,2 respectively.

輸入

3 1

輸出

-1

題解：

• 如果有解，那麼 n(n+1)/2 % n == k，因為長度為n的子區間是P本身，P的元素之和為n(n+1)/2.
• 假設k滿足上述條件，此時一定有解。如果是奇數，則k=0，則P={n, 1, n-1, 2, n-2, ..}，若為偶數，則P={n, n/2, 1, n-1, 2, n-2, ...}

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int n, k;
 5 
 6 void solve() {
 7     cin >> n >> k;
 8     int sum = n * (n+1) / 2;
 9     if (sum % n != k)   cout << "-1\n";
10     else {
11         if (n % 2 == 0) {
12             cout << n << " " << n / 2
 
;
13             for (int i = 1; i <= n/2 -1; i++) {
14                 cout << " " << i << " " << n - i;
15             }
16         }
17         else {
18             cout << n;
19             for (int i = 1; i <= n/2; i++) {
20                 cout << " " << i << "
 
 " << n - i;
21             }
22         }
23     }
24 }
25 
26 int main() {
27     solve();
28     cout << "\n";
29     return 0;
30 }