第一场 B Hints of sd0061

Posted 2020-09-30 Wally的博客

sd0061, the legend of Beihang University ACM-ICPC Team, retired last year leaving a group of noobs. Noobs have no idea how to deal with mm coming contests. sd0061 has left a set of hints for them. 

There are nn noobs in the team, the ii-th of which has a rating aiai. sd0061 prepares one hint for each contest. The hint for the jj-th contest is a number bjbj, which means that the noob with the (bj+1)(bj+1)-th lowest rating is ordained by sd0061 for the jj-th contest. 

The coach asks constroy to make a list of contestants. constroy looks into these hints and finds out: bi+bj≤bkbi+bj≤bk is satisfied if bi≠bj,bi≠bj, bi<bkbi<bk and bj<bkbj<bk. 

Now, you are in charge of making the list for constroy.

InputThere are multiple test cases (about 1010). 

For each test case: 

The first line contains five integers n,m,A,B,Cn,m,A,B,C. (1≤n≤107,1≤m≤100)(1≤n≤107,1≤m≤100) 

The second line contains mm integers, the ii-th of which is the number bibi of the ii-th hint. (0≤bi<n)(0≤bi<n) 

The nn noobs‘ ratings are obtained by calling following function nn times, the ii-th result of which is aiai. 

unsigned x = A, y = B, z = C;
unsigned rng61() {
  unsigned t;
  x ^= x << 16;
  x ^= x >> 5;
  x ^= x << 1;
  t = x;
  x = y;
  y = z;
  z = t ^ x ^ y;
  return z;
}

OutputFor each test case, output " Case #xx: y1y1 y2y2 ?? ymym" in one line (without quotes), where xx indicates the case number starting from 11 and yiyi (1≤i≤m)(1≤i≤m) denotes the rating of noob for the ii-th contest of corresponding case.Sample Input

3 3 1 1 1
0 1 2
2 2 2 2 2
1 1

Sample Output

Case #1: 1 1 202755
Case #2: 405510 405510

题目大意：使用题目中给出的生成函数，生成n个数，然后给出m个bi打印n个数中第bi+1大的数。

解题思路：对bi进行由小到大排序，然后从m到0使用nth_element进行查找，找出bi+1大的数

AC代码：