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：