# Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2421    Accepted Submission(s): 736

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

Input
There are multiple test cases (about
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 #" in one line (without quotes), where

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

Source

 1 #pragma comment(linker, "/STACK:102400000,102400000")
2 #include <bits/stdc++.h>
3 #include <cstdlib>
4 #include <cstdio>
5 #include <iostream>
6 #include <cstdlib>
7 #include <cstring>
8 #include <algorithm>
9 #include <cmath>
10 #include <cctype>
11 #include <map>
12 #include <set>
13 #include <queue>
14 #include <bitset>
15 #include <string>
16 #include <complex>
17 #define LL long long
18 #define mod 1000000007
19 using namespace std;
20 int n,m;
21 unsigned x,y,z,t,ans[10000007];
22 unsigned rng61(){
23     x^=x<<16;
24     x^=x>>5;
25     x^=x<<1;
26     t=x;
27     x=y;
28     y=z;
29     z=t^x^y;
30     return z;
31 }
32 unsigned  aa[10000007];
33 struct node
34 {
35     int xx;
36     int pos;
37     friend bool operator < (node aaa,node bbb)
38     {
39         return aaa.xx < bbb.xx;
40     }
41 }bb[105];
42 int main()
43 {
44     int t=0;
45     while(scanf("%d %d %u %u %u",&n,&m,&x,&y,&z)!=EOF){
46         for(int i=1; i<=m; i++){
47             scanf("%d",&bb[i].xx);
48             bb[i].pos=i;
49             }
50         for(int i=0; i<n; i++)
51             aa[i]=rng61();
52         sort(bb+1,bb+1+m);
53         bb[m+1].xx=n;
54         for(int i=m;i>=1;i--){
55             nth_element(aa,aa+bb[i].xx,aa+bb[i+1].xx);
56             ans[bb[i].pos]=aa[bb[i].xx];
57         }
58         printf("Case #%d:",++t);
59         for(int i=1;i<=m;i++)
60             printf(" %u",ans[i]);
61         printf("\n");
62     }
63     return 0;
64 }