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第一場 B Hints of sd0061

start star 代碼 acm left multipl 生成 ssi isf

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 mm coming contests. sd0061 has left a set of hints for them.

There are nn noobs in the team, the ii-th of which has a rating aiai. sd0061 prepares one hint for each contest. The hint for the
jj-th contest is a number bjbj, which means that the noob with the (bj+1)(bj+1)-th lowest rating is ordained by sd0061 for the jj-th contest.

The coach asks constroy to make a list of contestants. constroy looks into these hints and finds out: bi+bjbkbi+bj≤bk is satisfied if bibj,bi≠bj, bi<bkbi<bk and
bj<bkbj<bk.

Now, you are in charge of making the list for constroy.

InputThere are multiple test cases (about 1010).

For each test case:

The first line contains five integers n,m,A,B,Cn,m,A,B,C. (1n107,1m100)(1≤n≤107,1≤m≤100)

The second line contains mm integers, the ii-th of which is the number

bibi of the ii-th hint. (0bi<n)(0≤bi<n)

The nn noobs‘ ratings are obtained by calling following function nn times, the ii-th result of which is aiai.

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;
}

OutputFor each test case, output " Case #xx: y1y1 y2y2 ?? ymym" in one line (without quotes), where xx indicates the case number starting from 11 and yiyi (1im)(1≤i≤m) denotes the rating of noob for the ii-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

題目大意:使用題目中給出的生成函數,生成n個數,然後給出m個bi打印n個數中第bi+1大的數。

解題思路:對bi進行由小到大排序,然後從m到0使用nth_element進行查找,找出bi+1大的數

AC代碼:

第一場 B Hints of sd0061