HDU 6040 Hints of sd0061 stl

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2301    Accepted Submission(s): 689

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 jbj, 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

题意：算n次那个函数得到a数组  每次询问第b[i]大的

题解：预处理后对b数组排序然后用nth_element()就好了

因为bi+bj<=bk

所以下降速度很快

注意nth_element()第二个参数是n-1  n是第n大的

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;unsigned x,y,z;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;}unsigned a[10000007],ans[105];int b[105],pos[105];bool cmp(int i,int j){    return b[i]<b[j];}int main(){    int cas=1,i;    unsigned A,B,C;    int n,m;    while(scanf("%d%d%u%u%u",&n,&m,&A,&B,&C)!=EOF){        x=A;        y=B;        z=C;        for(i=1;i<=m;i++){            scanf("%d",&b[i]);    //        b[i]++;            pos[i]=i;        }        sort(pos+1,pos+1+m,cmp);        for(i=1;i<=n;i++)a[i]=rng61();        b[m+1]=n;        pos[m+1]=m+1;        printf("Case #%d:",cas++);        for(i=m;i>=1;i--){            nth_element(a+1,a+1+b[pos[i]],a+1+b[pos[i+1]]);            ans[pos[i]]=a[b[pos[i]]+1];        }        for(i=1;i<=m;i++)cout<<" "<<ans[i];        printf("\n");    }        return 0;}