HDU 6040 Hints of sd0061 —— 2017 Multi-University Training 1

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：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int MAXN=1e7+;
 unsigned n, m;
 unsigned rat[MAXN], b[MAXN],p[MAXN], a[MAXN];
 unsigned x,y,z;
 unsigned rng61() {
     unsigned t;
     x ^= x << ;
     x ^= x >> ;
     x ^= x << ;
     t = x;
     x = y;
     y = z;
     z = t ^ x ^ y;
     return z;
 }
 bool cmp(int s, int t)
 {
     return b[s]<b[t];
 }
 int main()
 {
     int k=;
     while(~scanf("%d %d %u %u %u", &n, &m, &x, &y, &z))
     {
         for(int i=;i<m;i++){
             p[i]=i;
             scanf("%d", b+i);
         }

         for(int i=;i<n;i++)
             rat[i]=rng61();
         sort(p, p+m,cmp);
         b[p[m]=m]=n;
         for(int i=m-;i>=;i--){
             if(b[p[i]]==b[p[i+]]){
                 a[p[i]]=a[p[i+]];
                 //continue;
             }
             nth_element(rat, rat+b[p[i]], rat+b[p[i+]]);
             a[p[i]]=rat[b[p[i]]];
         }
         printf("Case #%d:", ++k);
         for(int i=;i<m;i++)
             printf(" %u", a[i]);
         printf("\n");
     }
 }