HDU 6040---Hints of sd0061(STL)

题目链接

 

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 aisd0061 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+bjbk is satisfied if bibj, 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(1n107,1m100)

The second line contains m integers, the i-th of which is the number bi of the i-th hint. (0bi<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 #xy1 y2  ym" in one line (without quotes), where x indicates the case number starting from 1 and yi (1im) 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] ,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+bjbk is satisfied if bibjbi<bk and bj<bk;
 
思路:可以发现b[]数组不是很大,对于每个b[i]找a[]数组的第b[i]+1的数时,可以使用nth(a,a+p,a+n) 这个STL函数,进行一次o(n)的排序,使得a[p]之前的数均小于a[p],a[p]之后的数均大于a[p],所以a[p]即为我们要的数。在对每个b[]元素操作时,可以对b[]排序,从大到小进行计算,以为之前排序的右半部分不必再进行排序,这样会减少很多计算量。
 
代码如下:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1e7+5;
unsigned x,y,z, A,B,C;
unsigned a[N];
struct Node
{
    int x;
    int id;
    unsigned y;
}tr[105];
bool cmp(const Node s1,const Node s2)
{
    return s1.x<s2.x;
}
bool cmp2(const Node s1,const Node s2)
{
    return s1.id<s2.id;
}

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()
{
    ///cout << "Hello world!" << endl;
    int n,m,Case=1;
    while(scanf("%d%d%u%u%u",&n,&m,&A,&B,&C)!=EOF)
    {
        x = A, y = B, z = C;
        for(int i=0;i<n;i++)  a[i]=rng61();
        printf("Case #%d:",Case++);
        for(int i=1;i<=m;i++) scanf("%d",&tr[i].x),tr[i].id=i;
        sort(tr+1,tr+m+1,cmp);

        int p=n;
        for(int i=m;i>=1;i--)
        {
            int x = tr[i].x;
            nth_element(a,a+x,a+p);
            p=x;
            tr[i].y=a[x];
        }
        sort(tr+1,tr+m+1,cmp2);
        for(int i=1;i<=m;i++) printf(" %u",tr[i].y);
        puts("");
    }
    return 0;
}

 

posted @ 2017-07-28 15:45  茶飘香~  阅读(...)  评论(...编辑  收藏