题目描述
有两个长度都是N的序列A和B，在A和B中各取一个数相加可以得到N2N^2N2个和，求这N2N^2N2个和中最小的N个。
输入输出格式
输入格式：第一行一个正整数N；
第二行N个整数AiA_iAi, 满足Ai≤Ai+1A_ile A_{i+1}Ai≤Ai+1且Ai≤109A_ile 10^9Ai≤109;
第三行N个整数BiB_iBi, 满足Bi≤Bi+1B_ile B_{i+1}Bi≤Bi+1且Bi≤109B_ile 10^9Bi≤109.
【数据规模】
对于50%的数据中，满足1<=N<=1000；
对于100%的数据中，满足1<=N<=100000。
输出格式：输出仅一行，包含N个整数，从小到大输出这N个最小的和，相邻数字之间用空格隔开。
输入输出样例
输入样例#1：
复制
3 2 6 6 1 4 8
输出样例#1： 复制
3 6 7
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 700005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
int a[maxn], b[maxn];
map<pii, int>mp;
struct node {
int x, y;
bool operator<(const node&t)const {
if (a[x] + b[y] < a[t.x] + b[t.y])return false;
return true;
}
node(int X, int Y) { x = X, y = Y; }
};
priority_queue<node>q;
int main() {
//ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i++)rdint(a[i]);
for (int j = 1; j <= n; j++)rdint(b[j]);
q.push(node(1, 1));
for (int i = 1; i <= n; i++) {
while (mp[make_pair(q.top().x, q.top().y)])q.pop();
int tmpx = q.top().x, tmpy = q.top().y;
cout << a[tmpx] + b[tmpy] << ' ';
mp[make_pair(tmpx, tmpy)] = 1;
q.push(node(tmpx + 1, tmpy)); q.push(node(tmpx, tmpy + 1));
}
return 0;
}