2269:minval (priority queue heap sort)

2269:minval time limit: 3 Sec Memory limit:
Submissions: 638 Resolution: 65
Submitted State [Discussion Version] [Propositional person: external import] Title Description

There are two sequences A and b of length n, and the sum of each number in A and B can be obtained N2 and, the N2 and the smallest n.

Input

The first line enters a positive integer N (1<=n<=100000);

The second row n integer AI and ai<=109; the third row n integer bi and bi<=109.

Output

Output only one row, containing n integers, from small to large output this n smallest and, adjacent numbers separated by a space.

Sample input

51 3 2 4 56 3 4 1 7

Sample output

2 3 4) 4 5

/*Heap Sort (priority queue) * Maintain n minimum number * Don't forget to sort*/#include<queue>#include<iostream>#include<cstdio>#include<algorithm>using namespacestd;#defineMAXN 110000intN;intNUM1[MAXN], NUM2[MAXN]; Priority_queue<int>Min_num;intRESULT[MAXN];intMain () {CIN>>N;  for(intI=1; I<=n; i++) {scanf ("%d", &Num1[i]); }     for(intj=1; J<=n; J + +) {scanf ("%d", &Num2[j]); } sort (Num1+1, num1+n+1) ; Sort (num2+1, num2+1+N);  for(intI=1; I<=n; i++) {Min_num.push (num1[1] +Num2[i]); }     for(intI=2; I<=n; i++){         for(intj=1; J<=n; J + +){            if(Num1[i] + num2[j] <Min_num.top ())                 {Min_num.pop (); Min_num.push (Num1[i]+Num2[j]); }Else {                 Break ; }        }    }     for(intI=1; I<=n; i++) {Result[i]=Min_num.top ();     Min_num.pop (); }     for(intI=n; i>=1; i--) {printf ("%d", Result[i]); }    return 0 ; }

2269:minval (priority queue heap sort)