第一场 hdu 6040 Hints of sd0061

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 mm coming contests. sd0061 has left a set of hints for them. 

There are nn noobs in the team, the ii-th of which has a rating aiai. sd0061 prepares one hint for each contest. The hint for the jj-th contest is a number bjbj, which means that the noob with the (bj+1)(bj+1)-th lowest rating is ordained by sd0061 for the jj-th contest. 

The coach asks constroy to make a list of contestants. constroy looks into these hints and finds out: bi+bj≤bkbi+bj≤bk is satisfied if bi≠bj,bi≠bj, bi<bkbi<bk and bj<bkbj<bk. 

Now, you are in charge of making the list for constroy.

InputThere are multiple test cases (about 1010). 

For each test case: 

The first line contains five integers n,m,A,B,Cn,m,A,B,C. (1≤n≤107,1≤m≤100)(1≤n≤107,1≤m≤100) 

The second line contains mm integers, the ii-th of which is the number bibi of the ii-th hint. (0≤bi<n)(0≤bi<n) 

The nn noobs' ratings are obtained by calling following function nn times, the ii-th result of which is aiai. 

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;
}

OutputFor each test case, output " Case #xx: y1y1 y2y2 ⋯⋯ ymym" in one line (without quotes), where xx indicates the case number starting from 11 and yiyi (1≤i≤m)(1≤i≤m) denotes the rating of noob for the ii-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

题目大意：使用题目中给出的生成函数，生成n个数，然后给出m个bi打印n个数中第bi+1大的数。

解题思路：对bi进行由小到大排序，然后从m到0使用nth_element进行查找，找出bi+1大的数

AC代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
unsigned int a[10000005];
struct node
{
    int id,pos;
    unsigned int ans;
}cnt[105];
unsigned int x,y,z;
unsigned int rng61() {
  unsigned int t;
  x ^= x << 16;
  x ^= x >> 5;
  x ^= x << 1;
  t = x;
  x = y;
  y = z;
  z = t ^ x ^ y;
  return z;
}
int cmp1(const node &u,const node &v)
{
    if(u.pos!=v.pos)
    return u.pos<v.pos;
    else
    return u.id<v.id;
}
int cmp2(const node &u,const node &v)
{
    return u.id<v.id;
}
int main()
{
    int n,m,cas=0;
    //freopen("1008.in","r",stdin);
    //freopen("1008.out","w",stdout);
    while(~scanf("%d%d",&n,&m))
    {
        cas++;
        scanf("%u%u%u",&x,&y,&z);
        for(int i=0;i<n;i++)
        {
            a[i]=rng61();
            //printf("%d
",a[i]);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&cnt[i].pos);
            cnt[i].id=i;
        }
        sort(cnt,cnt+m,cmp1);
        cnt[m].pos=n;
        cnt[m].id=m;
//        for(int i=0;i<m;i++)
//        printf("%d %d
",cnt[i].pos,cnt[i].id);
        for(int i=m-1;i>=0;i--)
        {
            nth_element(a,a+cnt[i].pos,a+cnt[i+1].pos);
            //printf("%d
",a[cnt[i].pos]);
            //printf("%d
",a[cnt[i].pos]);
            cnt[cnt[i].id].ans=a[cnt[i].pos];
        }
        printf("Case #%d:",cas);
        for(int i=0;i<m;i++)
        {
            printf(" %u",cnt[i].ans);
        }
        printf("
");
    }
    return 0;
}