hdu 6040 Hints of sd0061【STL线性排序函数】

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1595 Accepted Submission(s): 466


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


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；了解这个函数后，本题基本就解决了

#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
const int max_n=1e7+10;
using namespace std; 
unsigned a[max_n],d[110];
unsigned x,y,z,n,m;

typedef pair<unsigned,int> P;
P s[110];

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()
{
	while(~scanf("%u %u %u %u %u",&n,&m,&x,&y,&z))
	{
		for(int i=0;i<n;i++)
			a[i]=rng61();
		for(int i=0;i<m;i++)
		{
			scanf("%u",&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--) //先大后小排 
		{
			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); //STL中的一种标准函数 
				d[s[i].second]=a[s[i].first];
			}
		}
		static int p=1;
		printf("Case #%d:",p++);  
        for(int i=0;i<m;i++) 
            printf(" %u",d[i]); 
        printf("\n"); 
	}
	return 0;
}