1245 the smallest N and

1245 the smallest n and

time limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingView Run ResultsTitle Description Description

There are two sequences A and b of length N, each taking a number in A and B can get n^2 and, the n^2 and the smallest n.

Enter a description Input Description

The first line enters a positive integer n, the second row n integers ai and ai≤10^9, and the third row n integer bi,
and bi≤10^9

Output description Output Description

Output only one row, containing n integers, from small to large output these n smallest and between adjacent numbers with
separated by spaces.

Sample input Sample Input

5

1 3 2) 4 5
6 3 4) 1 7

Sample output Sample Output

2 3 4) 4 5

Data range and Tips Data Size & Hint

"Data size" for 100% of data, meet 1≤n≤100000.

Category labels Tags Click here to expandHeap tree structure

Exercises

The most violent method:

We can figure out all the things and then sort them out, and obviously space and time will explode.

On-Line ideas:

(In fact, it is not very clear that the i*j-1 of the first n solution is the optimal solution)

Try to get rid of some of the impossible States.

First, or a A, a, or the same table:

.. ..

b\a	1	2	...	I	...	N
1
2
...
I
...
N

observed that for(I,JThis point, the element smaller than it has at leastIXJ−1 of them.
Since we are asking for the first n small, the point satisfying the requirement must satisfy at leastIxJ−1<n is ixj≤N.
So we can reduce the number of points to

⌊N1⌋+⌊N2⌋+...+⌊NI⌋+...+⌊NN⌋=O(N∑i= 1n1i) = o (nlog n)

Complexity of Time:O(nlog2n)
Space complexity:O(nlogn)

AC Code 1:

#include <cstdio>#include<algorithm>using namespacestd;#defineN 100010intn,cnt,a[n],b[n],c[n* -];intMain () {scanf ("%d",&N);  for(intI=1; i<=n;i++) scanf ("%d", A +i);  for(intI=1; i<=n;i++) scanf ("%d", B +i); Sort (a+1, a+n+1); Sort (b+1, b+n+1);  for(intI=1; i<=n;i++){         for(intj=1; i*j<=n;j++) {c[++cnt]=a[i]+B[j]; }} sort (c+1, c+cnt+1);  for(intI=1; i<=n;i++) printf ("%d", C[i]); return 0;}

AC Code 2:

#include <cstdio> #include <algorithm> #include <queue> #include <vector>using namespace std;# Define N 100010int N,cnt,a[n],b[n];p riority_queue<int,vector<int>,greater<int> >que;int Main () {    scanf ("%d", &n);    for (int i=1;i<=n;i++) scanf ("%d", a+i);    for (int i=1;i<=n;i++) scanf ("%d", b+i);    Sort (a+1,a+n+1);    Sort (b+1,b+n+1);    for (int i=1;i<=n;i++) {for        (int j=1;i*j<=n;j++) {            que.push (a[i]+b[j]);         }    }    for (int i=1;i<=n;i++) {        printf ("%d", que.top ()); Que.pop ();    }    return 0;}

1245 the smallest N and