Zhengzhou Light Industry OJ minval

minval

Topic description

There are two sequences A and B of length N. If you add any number in A and B, you can get N2 sums, and find the smallest N of these N2 sums.

enter

Enter a positive integer N (1<=N<=100000) in the first line;

The second row contains N integers Ai and Ai<=109; the third row contains N integers Bi and Bi<=109.

output

The output is only one line, containing n integers, the n smallest sums are output from small to large, and adjacent numbers are separated by spaces.

sample input

51 3 2 4 56 3 4 1 7

Sample output

2 3 4 4 5

source

 priority queue

#include<stdio.h>
#include<queue>
#include<algorithm>
using namespace std;
long long a[100005],b[100005],ans[100005];
int main(){
	int N;
	//priority_queue<long long,vector<long long>,greater<long long> >q;
	priority_queue<long long>q;//The default sorting from small to large
	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++)
		q.push(a[1]+b[i]);
	for(int i=2;i<=N;i++){
		for(int j=1;j<=N;j++){
			if(a[i]+b[j]<q.top()){
				q.pop();
				q.push(a[i]+b[j]);
			}
			else
			break;//If the current value is greater than all the values ​​in the queue (because it is a priority queue, as long as it is greater than one value, it is greater than all values), jump out,
					//And because the array is sorted from small to large, if the current value is large, the subsequent value will be larger 
		}
	}
	int k=1;
	while(!q.empty()){//Because the queue is placed from large to small by default, so the output from small to large needs to add an array 
		ans[k++]=q.top();
		q.pop();
	}
	for(int i=N;i>=1;i--){
		if(i==1)
		printf("%lld\n",ans[i]);
		else
		printf("%lld ",ans[i]);
	}
	return 0;
}