HDU6040-Hints_of_sd0061-STL/快排思想/线性时间找第k大

Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1638 Accepted Submission(s): 479

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

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

Source

2017 Multi-University Training Contest - Team 1

题目真的难读……

大意就是给你递推式和递推初始条件，对应n个数字，要你回答m个“第bi小的数是谁”问题

因为数据量达到了

107

，排个序就直接T了，所以不能这样做

这里运用的就是快排时候的一个性质：快排时只需要知道这个数是在数列中哪个位置就好，那么考虑在已经知道这个数为n，对于任意小于n的询问，我们就没必要管比n大的数，也就是说我们排序的范围就减小了

同时运用快排的方式随机的找第n小元素时间复杂度是O(n)，而且每次询问时的n都不增，这样时间也就是可以接受的

要么手写找第k大，要么就用STL中的nth_element()方法

AC代码

#include <iostream>#include <queue>#include <algorithm>#include <cstring>#include <string>#define LL long long#define ULL unsigned long long#define mem(a,n) memset(a,n,sizeof(a))#define fread freopen("in.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)#define N 10001000#define INF 0x3f3f3f3f#define eps 1e-9using namespace std;unsigned x,y,z,num[N],ans[110];int query[110],index[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;}inline bool cmp(const int &idx1,const int &idx2){    return query[idx1]<query[idx2];}int main(){    ios::sync_with_stdio(false);    int n,m,A,B,C;    int kase=0;    while(cin>>n>>m>>A>>B>>C){        cout<<"Case #"<<++kase<<": ";        x=A,y=B,z=C;        for(int i=0;i<n;++i){            num[i]=rng61();        }        for(int i=0;i<m;++i){            cin>>query[i];            index[i]=i;        }        sort(index,index+m,cmp);        query[index[m]=m]=n;        for(int i=m-1;~i;--i){            if(index[i]==index[i+1]){                ans[index[i]]=ans[index[i+1]];                continue;            }            nth_element(num,num+query[index[i]],num+query[index[i+1]]);            ans[index[i]]=num[query[index[i]]];        }        for(int i=0;i<m;++i){            cout<<ans[i]<<(i==m-1)[" \n"];        }    }    return 0;}