AtCoder Grand Contest 007 B - Construct Sequences

固定a[p[i]] + b[p[i]]和b[i]
令a[p[i]] + b[p[i]]为1, 2, 3...，b[i]为-1 * n, -2 * n, -3 * n...(计算出a[i]，然后再考虑偏移)
时间复杂度：O(n)
空间复杂度：O(n)

  1. static final int N = 20010;
  2.     static int[] a = new int[N], b = new int[N];
  3.     static int[] p = new int[N];
  4.     static int[] x = new int[N];
  5.     static int n;
  6.     static void solve() {
  7.         n = ni();
  8.         for (int i = 1; i <= n; i++)
  9.             p[i] = ni();
  10.         for (int i = 1; i <= n; i++) {
  11.             x[p[i]] = i;
  12.             b[i] = n * (-i);
  13.         }
  15.         for (int i = 1; i <= n; i++) {
  16.             a[i] = x[i] - b[i];
  17.             b[i] += (n + 1) * n;
  18.         }
  20.         for (int i = 1; i <= n; i++) {
  21.             out.print(a[i] + " ");
  22.         }
  23.         out.println();
  24.         for (int i = 1; i <= n; i++) {
  25.             out.print(b[i] + " ");
  26.         }
  28.     }