2017 Multi-University Training Contest - Team 1 ：Hints of sd0061（hdu 6040）

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1540    Accepted Submission(s): 453

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

 
//题意：输入n,m,A,B,C，n代表要循环题目给的函数n次，每次函数结束的值都储存在Y[]（自己定义的一个数组），m表示下面那排有m个数，A,B,C分别为题目给的函数里x,y,z的初值。第二排m个数表示输出Y[]中第那个数大的值是多少。

//思路：这题要卡时间，排序不能用sort，会TLE，要用nth_element(start,start+n,end)这个函数，这个函数在c++STL库中，包含于<algorithm>，作用是把一个数组里第n大的元素放在第n个位置，时间复杂度O(nlogn)，和快排（sort）是一样的，但注意，不用把整个数组都排序完，具体怎么用自行百度。

还有一点是你输入的m个要查找的位置要先从大到小排序（用sort ，因为m非常小），尽量人为降低nth_element()的复杂度，不然会TLE，但注意输出的时候要按它m个数输入的顺序！

还有就是unsigned类型的输入是%u 。

//多校，被高中生血虐的辛酸史[捂脸]...

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 1e7 + 100;
const int MAX = 100 + 10;

typedef struct {
unsigned id;
unsigned val;
}rrank;

unsigned x, A, y, B, z, C;
unsigned Y
;
rrank Rank[MAX];
unsigned ans[MAX];

//这里从小到大排序了
//是因为下面的循环我是从m-1 -> 0
//相当于还是从大到小排序
bool cmp(rrank p1, rrank p2)
{
return p1.val < p2.val;
}

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 Case = 1;
int n, m;
while (scanf("%u%u%u%u%u", &n, &m, &A, &B, &C) != EOF)
{
x = A;
y = B;
z = C;
for (int i = 0; i < m; i++)
{
scanf("%u", &Rank[i].val);
Rank[i].id = i;
}

sort(Rank, Rank + m, cmp);
Rank[m].val = n;
Rank[m].id = m;
printf("Case #%d: ", Case++);

for (int i = 0; i < n; i++)
{
Y[i] = rng61();
}

for (int i = m - 1; i >= 0; i--)
{
if (Rank[i].val == Rank[i + 1].val)
ans[Rank[i].id] = ans[Rank[i + 1].id];
else
{
nth_element(Y, Y + Rank[i].val, Y + Rank[i + 1].val);
ans[Rank[i].id] = Y[Rank[i].val];
}
}

for (int i = 0; i < m - 1; i++)
{
printf("%u ", ans[i]);
}
printf("%u\n", ans[m - 1]);

}
return 0;
}