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

2024-11-13 10:25:01   202人阅读

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<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 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]<b[t];
24 }
25 int main()
26 {
27     int k=0;
28     while(~scanf("%d %d %u %u %u", &n, &m, &x, &y, &z))
29     {
30         for(int i=0;i<m;i++){
31             p[i]=i;
32             scanf("%d", b+i);
33         }
34             
35         for(int i=0;i<n;i++)
36             rat[i]=rng61();
37         sort(p, p+m,cmp);
38         b[p[m]=m]=n;
39         for(int i=m-1;i>=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<m;i++)
49             printf(" %u", a[i]);
50         printf("\n");
51     }
52 }

 

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


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