hdu 6040 Hints of sd0061(stl: nth_element(arr,arr+k,arr+n))
Hints of sd0061
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2262 Accepted Submission(s): 673
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.
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.
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
Recommend
题意:给你一个长度为n的序列,序列由题面给的函数生成。然后m次询问,询问这个序列上第bi小的数。
解题思路:新学习了一个STL,C++中的nth_element(arr,arr+k,arr+n),将长度为n的数组arr进行划分,第k-1位置上就是第k大的数(下标从0开始算),这个函数近似线性,在找到第k大的时候,前k-1个数均是小于arr[k]的,因为输入保证任意两个小的之和小于第三个,所以查询数列的间隔一定大于等于斐波那契,所以从大到小查询的话,每次至少能去掉一半的区间,根据这个可以减少搜索量
hdu上要用G++交,C++超时
#include <iostream> #include<cstdio> #include<functional> #include<cstring> #include<algorithm> using namespace std; unsigned a[10000005]; struct node { int k,id; }b[105]; int n,m; 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; } bool cmp(node a,node b) { return a.k>b.k; } bool cmp2(node a,node b) { return a.id<b.id; } int main() { int cas=0; while(~scanf("%d%d%u%u%u",&n,&m,&x,&y,&z)) { for(int i=1;i<=m;i++) { scanf("%d",&b[i].k); b[i].id=i; } for(int i=0;i<n;i++) a[i]=rng61(); sort(b+1,b+1+m,cmp); b[0].k=n; for(int i=1;i<=m;i++) nth_element(a,a+b[i].k,a+b[i-1].k); sort(b+1,b+1+m,cmp2); printf("Case #%d:",++cas); for(int i=1;i<=m;i++) printf(" %u",a[b[i].k]); printf("\n"); } return 0; }