Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 2297    Accepted Submission(s): 687

Problem Description

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 

m coming contests. sd0061 has left a set of hints for them.

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

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

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

 

Input

There are multiple test cases (about 

10).

For each test case:

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

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

The n noobs' ratings are obtained by calling following function n times, the i-th result of which is ai.

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; }

Output

For each test case, output "

Case #x: y1 y2 ⋯ ym" in one line (without quotes), where 

x indicates the case number starting from 1 and yi (1≤i≤m) denotes the rating of noob for the i-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 40551

 

 

题目大意：用题目所给的程序生成a数组，m个询问，每个询问输出a从小至大排序后第bi个数。

思路：按照题意进行排序，不过输出ai前用sort会超时，用nth_element()可以避免TLE。

 

AC代码：

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 using namespace std; 6 const int MAXN=1e7+5; 7 unsigned n, m; 8 unsigned rat[MAXN], b[MAXN],p[MAXN], a[MAXN]; 9 unsigned x,y,z;10 unsigned rng61() {11     unsigned t;12     x ^= x << 16;13     x ^= x >> 5;14     x ^= x << 1;15     t = x;16     x = y;17     y = z;18     z = t ^ x ^ y;19     return z;20 }21 bool cmp(int s, int t)22 {23     return b[s]
           
            =0;i--){40             if(b[p[i]]==b[p[i+1]]){41                 a[p[i]]=a[p[i+1]];42                 //continue;43             } 44             nth_element(rat, rat+b[p[i]], rat+b[p[i+1]]);45             a[p[i]]=rat[b[p[i]]];46         }47         printf("Case #%d:", ++k);48         for(int i=0;i