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
题意：输入n,m,A,B,C 由ABC和题中提供的函数可以 得到n个数的数组a[n](就是for循环跑n此题给函数就可以了) ，然后输入m个数 表示 数组b[m]，现在对于每个b[i]输出a[]数组中的第b[i]+1小的数。题中给出b[]数组的限制bi+bj≤bk is satisfied if bi≠bj, bi<bk and bj<bk；
思路：利用nth_element(a,a+k,a+n)函数，其原理类似于快拍，k项之前的都小于a[k](不一定有序)，k后面的都是大于a[k]，复杂度O(n)。
1 #include <iostream>
2 #include <algorithm>
3 #include <cstdio>
4 #include <cstring>
5 using namespace std;
6 const int N=1e7+5;
7 unsigned x,y,z, A,B,C;
8 unsigned a[N];
9 struct Node
10 {
11 int x;
12 int id;
13 unsigned y;
14 }tr[105];
15 bool cmp(const Node s1,const Node s2)
16 {
17 return s1.x<s2.x;
18 }
19 bool cmp2(const Node s1,const Node s2)
20 {
21 return s1.id<s2.id;
22 }
23
24 unsigned rng61() {
25 unsigned t;
26 x ^= x << 16;
27 x ^= x >> 5;
28 x ^= x << 1;
29 t = x;
30 x = y;
31 y = z;
32 z = t ^ x ^ y;
33 return z;
34 }
35
36 int main()
37 {
38 ///cout << "Hello world!" << endl;
39 int n,m,Case=1;
40 while(scanf("%d%d%u%u%u",&n,&m,&A,&B,&C)!=EOF)
41 {
42 x = A, y = B, z = C;
43 for(int i=0;i<n;i++) a[i]=rng61();
44 printf("Case #%d:",Case++);
45 for(int i=1;i<=m;i++) scanf("%d",&tr[i].x),tr[i].id=i;
46 sort(tr+1,tr+m+1,cmp);
47
48 int p=n;
49 for(int i=m;i>=1;i--)
50 {
51 int x = tr[i].x;
52 nth_element(a,a+x,a+p);
53 p=x;
54 tr[i].y=a[x];
55 }
56 sort(tr+1,tr+m+1,cmp2);
57 for(int i=1;i<=m;i++) printf(" %u",tr[i].y);
58 puts("");
59 }
60 return 0;
61 }//转自https://www.cnblogs.com/chen9510/p/7250651.html
原文地址：https://www.cnblogs.com/llllrj/p/9424835.html