CodeForcesdiv1：995C - Leaving the Bar（随机算法+贪心）

For a vector →v=(x,y)v→=(x,y), define |v|=√x2+y2|v|=x2+y2.

Allen had a bit too much to drink at the bar, which is at the origin. There are nn vectors →v1,→v2,⋯,→vnv1→,v2→,⋯,vn→. Allen will make nn moves. As Allen's sense of direction is impaired, during the ii-th move he will either move in the direction →vivi→ or −→vi−vi→. In other words, if his position is currently p=(x,y)p=(x,y), he will either move to p+→vip+vi→ or p−→vip−vi→.

Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position pp satisfies |p|≤1.5⋅106|p|≤1.5⋅106 so that he can stay safe.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of moves.

Each of the following lines contains two space-separated integers xixi and yiyi, meaning that →vi=(xi,yi)vi→=(xi,yi). We have that |vi|≤106|vi|≤106 for all ii.

Output

Output a single line containing nn integers c1,c2,⋯,cnc1,c2,⋯,cn, each of which is either 11 or −1−1. Your solution is correct if the value of p=∑ni=1ci→vip=∑i=1ncivi→, satisfies |p|≤1.5⋅106|p|≤1.5⋅106.

It can be shown that a solution always exists under the given constraints.

Examples

input

Copy

3
999999 0
0 999999
999999 0

output

Copy

1 1 -1 

input

Copy

1
-824590 246031

output

Copy

1 

input

Copy

8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899

output

Copy

1 1 1 1 1 1 1 -1 

题意：给定一些向量，你可以改变它的符号，使得这些向量之和的长度小于1.5e6。

思路：考虑到每条边的长度小于1e6，所以任意两条边的向量和长度小于1.414e6<1.5e6。任意三条边的向量和（可以改变方向）的长度也成立，下面给出证明...

所以我们贪心, 保证向量和离原点更近. 然后下面的代码AC了,然后又被FST了.是因为每次我贪心原则是一样的.最后的结果有可能大于1.5e6.  我们需要加一些随机性, 多次贪心,直到结果满足题意.（这里随机的作用就算多次贪心，直到满足）

证明：每三个向量abc中都能找到两个向量合起来 <= 1e6，然后不断合并，最后只会剩下一个或者两个向量。

对于a和b，1，若ab小于60度，其中一个转向，则变为大于120度，<1e6；2，ab角度大于120度，则其向量和的长度小于1e6。3，角度在60到120度之间，那么第三边c，不论怎么放，都或与a或与b满足上面的关系，使得其向量和长度小于1e6，那么一直这样合并，最后只剩下一边或者两边，而两边的情况最坏为1.414e6

FST代码,WA79:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=;
const ll p=;
ll ans[maxn+],x[maxn+],y[maxn+];
ll X,Y,P;
int main()
{
    int N,i,j; P=p*p;scanf("%d",&N);
    for(i=;i<=N;i++){
       scanf("%I64d%I64d",&x[i],&y[i]);
       if(X>=&&Y>=&&x[i]<=&&y[i]<=) X+=x[i],Y+=y[i],ans[i]=;
       else if(X>=&&Y>=&&x[i]>=&&y[i]>=) X-=x[i],Y-=y[i],ans[i]=-;
       else if(X<=&&Y<=&&x[i]<=&&y[i]<=) X-=x[i],Y-=y[i],ans[i]=-;
       else if(X<=&&Y<=&&x[i]>=&&y[i]>=) X+=x[i],Y+=y[i],ans[i]=;
       else if((X+x[i])*(X+x[i])+(Y+y[i])*(Y+y[i])<(X-x[i])*(X-x[i])+(Y-y[i])*(Y-y[i])) X+=x[i],Y+=y[i],ans[i]=;
       else X-=x[i],Y-=y[i],ans[i]=-;
    }
    for(i=;i<=N;i++) printf("%d ",ans[i]);
    return ;
}

AC代码:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=;
const ll p=;
ll X,Y,P; int ans[maxn];
struct in{ ll x,y,id; }s[maxn];
int main()
{
    int N,i,j; P=p*p;scanf("%d",&N);
    for(i=;i<=N;i++){
       scanf("%I64d%I64d",&s[i].x,&s[i].y); s[i].id=i;
    }
    while(true){
       random_shuffle(s+,s+N+); X=Y=;
       for(i=;i<=N;i++){
          if((X+s[i].x)*(X+s[i].x)+(Y+s[i].y)*(Y+s[i].y)>(X-s[i].x)*(X-s[i].x)+(Y-s[i].y)*(Y-s[i].y)) X-=s[i].x,Y-=s[i].y,ans[s[i].id]=-;
          else X+=s[i].x,Y+=s[i].y,ans[s[i].id]=;
       }
       if(X*X+Y*Y<=P) {
               for(i=;i<=N;i++) printf("%d ",ans[i]); return ;
       }
    }
    return ;
}

（div1E也是需要随机算法！！！