2017多校联合一／hdu6040Hints of sd0061

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2311    Accepted Submission(s): 693

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

 

Source

2017 Multi-University Training Contest - Team 1

本题注意一个接近线性的查找函数，STL中的nth_element()方法的使用 通过调用nth_element(start, start+n, end) 方法可以使第n大元素处于第n位置（从0开始,其位置是下标为 n的元素），并且比这个元素小的元素都排在这个元素之前，比这个元素大的元素都排在这个元素之后，但不能保证他们是有序的；

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;using namespace std;//hdu 6040 m次线性查找第K小// nth_elememnt(A,A+k,A+n)const int maxn=1e7+10;int cas=1;unsigned n,m,x,y,z;unsigned a[maxn],d[105];pair<unsigned,int>s[105];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 main(int argc, const char * argv[]){    while(cin>>n>>m>>x>>y>>z)    {        for(int i=0;i<n;i++)            a[i]=rng61();        for(int i=0;i<m;i++)        {            cin>>s[i].first;            s[i].second=i;        }        sort(s,s+m);        s[m].first=n;        s[m].second=m;        for(int i=m-1;i>=0;i--)        {            if(s[i].first==s[i+1].first) d[s[i].second]=d[s[i+1].second];            else            {                nth_element(a,a+s[i].first,a+s[i+1].first);                d[s[i].second]=a[s[i].first];            }        }        cout<<"Case #"<<cas++<<":";        for(int i=0;i<m;i++)            cout<<" "<<d[i];        cout<<endl;    }    return 0;}