2017 Multi-University Training Contest

根据题意卡常数

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
int n,m;
unsigned A,B,C;
unsigned x,y,z;
map <int,unsigned> M;
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[11234567];
struct querry
{
    int id;
    int q;
    unsigned ans;
}b[200];
bool cmpq(querry x,querry y)
{
    return x.q > y.q;
}
bool cmpid(querry x,querry y)
{
    return x.id < y.id;
}
int  part(int low, int  high)
{
    swap(a[high],a[(low+high)/2]);
    unsigned pivot = a[high];
    while(low < high){
        while(low < high&&a[low] <= pivot) low++;
        a[high] = a[low];
        while(low < high&&a[high] >= pivot) high--;
        a[low] = a[high];
    }
    a[high]=pivot;
    return high;
}
unsigned quicksort(int l, int  r, int  k){
    int  pos = part(l,r);
    if(pos  == k) return a[pos];
    else if(pos > k) return quicksort(l,pos-1,k);
    else return quicksort(pos+1,r,k);
}
int main()
{
    int cas = 0;
    while (~scanf("%d%d%u%u%u",&n,&m,&A,&B,&C))
    {
        M.clear();
        x = A, y = B, z = C;
        for (int i = 1; i<=n; i++)
        {
            a[i] = rng61();
        }
        for (int i = 1; i<=m; i++)
        {
            scanf("%d",&b[i].q);
            b[i].id = i;
        }
        sort(b+1,b+m+1,cmpq);
        int last = n;
        for (int i = 1; i <= m;i++)
        {
            if (M.find(b[i].q+1)!=M.end()) b[i].ans = M[b[i].q+1];
            else b[i].ans = quicksort(1,last,b[i].q+1);
            M[b[i].q+1] = b[i].ans;
            last = b[i].q+1;
        }
        sort(b+1,b+m+1,cmpid);
        printf("Case #%d:",++cas);
        for (int i = 1; i<=m;i++)
            printf(" %u",b[i].ans) ;
        puts("");
    }
    return 0;
}

从最大的取值到最小的取值依次使用近似线性复杂度的求第 kk 小的方法即可，该方法的思想与快排相似，可以保证前 k - 1k−1 小的元素都放置在第 kk 小元素的前面，这样枚举的时候就可以依次减少每次的枚举量。

http://bestcoder.hdu.edu.cn/blog/