hdu 6040 Hints of sd0061(nth_element)

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1839    Accepted Submission(s): 539

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 10 1 22 2 2 2 21 1

 

Sample Output

Case #1: 1 1 202755Case #2: 405510 405510

解题报告

C++ STL中有一个库函数，nth_element。关于它的用法如下：

    有了这个函数，我们只需要离线的对查询排序，然后对每个查询的值在不断分割的子区间上调用这个函数（实际上就和快速排序的实现差不多），这样就可有很少冗余度的得到我们所需要的答案。

    这样写复杂度的显然是远小于快排的O(N * log N)，详细的复杂度计算引用官方题解：

#include <algorithm>#include <string.h>#include <iostream>#include <stdio.h>#include <string>#include <vector>#include <queue>#include <map>#include <set>#include <stack>using namespace std;typedef long long LL;const int N = 1e7+10;unsigned a[N],ans[110];int b[110], pos[110] ;const LL mod = 1e9+7;int A, B, C;unsigned x, y, z;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;}int cmp(int p,int q){return b[p]<b[q];}int main(){    int t, ncase=1;    int n, m;    while(scanf("%d %d %d %d %d", &n, &m, &A, &B, &C)!=EOF)    {        x=A,y=B,z=C;        for(int i=0;i<n;i++)  a[i]=rng61();        for(int i=0;i<m;i++) scanf("%d", &b[i]),pos[i]=i;        sort(pos,pos+m,cmp);        b[m]=n,pos[m]=m;        for(int i=m-1;i>=0;i--)        {            if(b[pos[i]]==b[pos[i+1]])  ans[pos[i]]=ans[pos[i+1]];            else            {                nth_element(a,a+b[pos[i]],a+b[pos[i+1]]);                ans[pos[i]]=a[b[pos[i]]];            }        }        printf("Case #%d:",ncase++);        for(int i=0;i<m;i++)  printf(" %u",ans[i]);        cout<<endl;    }    return 0;}