HDU 6040 stl

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 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 405510

 

Source

2017 Multi-University Training Contest - Team 1

 

#pragma comment(linker, "/STACK:102400000,102400000")
#include <bits/stdc++.h>
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define LL long long
#define mod 1000000007
using namespace std;
int n,m;
unsigned x,y,z,t,ans[10000007];
unsigned rng61(){
    x^=x<<16;
    x^=x>>5;
    x^=x<<1;
    t=x;
    x=y;
    y=z;
    z=t^x^y;
    return z;
}
unsigned  aa[10000007];
struct node
{
    int xx;
    int pos;
    friend bool operator < (node aaa,node bbb)
    {
        return aaa.xx < bbb.xx;
    }
}bb[105];
int main()
{
    int t=0;
    while(scanf("%d %d %u %u %u",&n,&m,&x,&y,&z)!=EOF){
        for(int i=1; i<=m; i++){
            scanf("%d",&bb[i].xx);
            bb[i].pos=i;
            }
        for(int i=0; i<n; i++)
            aa[i]=rng61();
        sort(bb+1,bb+1+m);
        bb[m+1].xx=n;
        for(int i=m;i>=1;i--){
            nth_element(aa,aa+bb[i].xx,aa+bb[i+1].xx);
            ans[bb[i].pos]=aa[bb[i].xx];
        }
        printf("Case #%d:",++t);
        for(int i=1;i<=m;i++)
            printf(" %u",ans[i]);
        printf("
");
    }
    return 0;
}