1245 min N sum, 1245 min

1245 min N sum, 1245 min
1245 minimum N sum

 

Time Limit: 1 s space limit: 128000 KB title level: Diamond Title Description Description

There are two sequences A and B with the length of N. For each number of A and B, we can get N ^ 2 and calculate the N ^ 2 and the N smallest N.

Input description Input Description

Enter a positive integer N in the first line; N integers Ai In the second line and Ai ≤ 10 ^ 9; N integers in the third line Bi,
And Bi ≤ 10 ^ 9

Output description Output Description

The output contains only one row and n integers. The smallest sum of N is output from small to large.
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 prompt Data Size & Hint

[Data scale] for 100% of data, 1 ≤ N ≤ 100000.

 1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 int a[100001]; 7 int b[100001]; 8 priority_queue<int>ans; 9 int can[1001][1001];10 int c[100001];11 int main()12 {13     int n;14     scanf("%d",&n);15     for(int i=1;i<=n;i++)16         scanf("%d",&a[i]);17     for(int i=1;i<=n;i++)18         scanf("%d",&b[i]);19     sort(a+1,a+n+1);20     sort(b+1,b+n+1);21     int now=1;22     for(int i=1;i<=n;i++)23     {24         ans.push(a[1]+b[i]);25     }26     for(int i=2;i<=n;i++)27     {28         for(int j=1;j<=n;j++)29         {30             int sum=a[i]+b[j];31             if(sum>=ans.top())32             {33                 break;34             }35             else36             {37                 ans.pop();38                 ans.push(sum);39             }40         }41     }42     //now=1;43     for(int i=1;i<=n;i++)44     {45         c[i]=ans.top();46         ans.pop();47     }48     for(int i=n;i>=1;i--)49     printf("%d ",c[i]);50     return 0;51 }

That is, no heap is used, and priority queue is usually used =. =