hdu1329 Ching and Magic

Address: http://acm.uestc.edu.cn/#/problem/show/1329

Topic:

Ching and MagicTime limit:1200/800ms (java/others) Memory limit:65535/65535kb (java/others)SubmitStatus

"Your membrane will not save you"--clam

In the way to save the princess, the Qing school elder sister, the blade has long been rusty stains.

A day Ching is worrying about the problem of weapons, met the tree in the days of the line Liao.

Day Line Liao Mouth Micro-Yang, seems to see through the Ching of the mind, deliberately wait here.

"Young man, do you desire to master the power of thunder?" "The day Liao is asking."

Was almost a salted fish Ching gladly agreed. Yu Shiqing began to follow the Magic Master Day Liao learning the power of magic.

Just getting started, Ching found that each magic is made up of two basic elements, A and B elements.

And the magic of each magic is the synthesis of the magic of A and B elements of the size and.

For example, a element with a size of 3 and a B element with a size of 6 can form a magical magic of 9.

Now the Qing school sister collected N "> N n a element and N "> n n a B element.

Keen Ching immediately found that he could assemble N & #x2217; N " >n∗n n∗n magic.

Modest Ching doesn't want to jump too much, so he's ready to N & #x2217; N " >n∗n n∗n magic of the smallest N "> n n species show to the day Liao inspection.

Now Ching want to know what the smallest n-n species of n ∗nn∗n Magic is.

Of course, from small to large output OH ~

Input

First line an integer N "> N n

The next line has N "> N n number, indicating N "> n n a element

The next line has N "> N n number, indicating N "> n n a b element

1 & #x2264; N & #x2264; 100000 " >1≤n≤100000 1≤n≤100000

1≤A[i],B[i]≤1000000000 1≤a[i],b[i]≤1000000000

Output

Output N ">< Span id= "mathjax-span-242" class= "Mrow" >n n line, one integer per line

Represents the smallest n n of n ∗nn∗n Magic

Sample Input and output

Sample Input	Sample Output
51 3 2 4 5 6 3 4 1 7	23445

Ideas:

is to first drag a[i]+b[0] into a priority queue, and then remove the minimum value a[i]+b[k] at the same time, and throw a[i]+b[k+1] in, and then until the n number is removed.

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cmath>5#include <cstring>6#include <queue>7#include <stack>8#include <map>9#include <vector>Ten#include <cstdlib> One#include <string> A  - #definePI ACOs ((double)-1) - #defineE exp (double (1)) the using namespacestd; - inta[100000],b[100000]; - intc[100000]; -  + intMain (void) - { +     intN; ACin>>N; at      for(intI=1; i<=n;i++) -scanf"%d",&a[i]); -Sort (A +1, a+n+1); -      for(intI=1; i<=n;i++) -scanf"%d",&b[i]); -Sort (b +1, b+n+1); inpriority_queue<pair<int,int> >s; -      for(intI=1; i<=n;i++) toS.push (Make_pair (a[i]+b[1]),1)); +      for(intI=1; i<=n;i++) -     { thepair<int,int> temp=s.top (); S.pop (); *c[i]=-Temp.first; $         intk=Temp.second;Panax Notoginseng         if(temp.second<n) S.push (Make_pair (-(-temp.first-b[k]+b[k+1]), K +1)); -     } the      for(intI=1; i<=n;i++) +printf"%d\n", C[i]); A     return 0; the } +  -  $ //struct Node $ //{ - //int sum,b; - //node (int a,int b): Sum (a), B (b) {} the //bool operator< (node &a) const - //    {Wuyi //return sum>a.sum; the //    } - //};

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hdu1329 Ching and Magic