codeforces 101C C. Vectors(数学)

题目链接：

C. Vectors

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:

• Turn the vector by 90 degrees clockwise.
• Add to the vector a certain vector C.

Operations could be performed in any order any number of times.

Can Gerald cope with the task?

Input

The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).

Output

Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).

Examples

input

0 0
1 1
0 1

output

YES

input

0 0
1 1
1 1

output

YES

input

0 0
1 1
2 2

output

NO
 
题意：
 
两个操作,1把A向量旋转90度,2把C向量加到A向量上,现在两个操作可以执行任意多次,顺序也是任意的,问能否得到B向量;
 
思路：
 
假设D为C旋转90度的向量,E为A旋转后的向量;
可以发现最后得到的都是a*C+b*D=B-E;
这就变成了一个判断一个二元一次方程组是否有整数解的问题;
那么就判断好了;
 
AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=(1<<20)+10;
const int maxn=1e6+10;
const double eps=1e-12;
 
LL ax,ay,bx,by,cx,cy;
int solve(LL x,LL y)
{
    x=bx-x;y=by-y;
    if(cx==0&&cy==0)
    {
        if(x==0&&y==0)return 1;
        return 0;
    }
    else if(cx==0)
    {
        LL g=x/cy,h=y/cy;
        if(g*cy==x&&h*cy==y)return 1;
        return 0;
    }
    else if(cy==0)
    {
        LL g=x/cx,h=y/cx;
        if(g*cx==x&&h*cx==y)return 1;
        return 0;
    }
    else
    {
        LL tx=x*cy,ty=y*cx;
        LL b=(tx-ty)/(cy*cy+cx*cx);
        if(b*(cx*cx+cy*cy)!=(tx-ty))return 0;
        else
        {
            LL a=(x-b*cy)/cx;
            if(a*cx+b*cy==x)return 1;
            return 0;
        }
    }
}
int main()
{
    read(ax);read(ay);
    read(bx);read(by);
    read(cx);read(cy);
    int f=0;
    if(solve(ax,ay))f=1;
    if(solve(-ax,-ay))f=1;
    if(solve(ay,-ax))f=1;
    if(solve(-ay,ax))f=1;
    if(f)cout<<"YES\n";
    else cout<<"NO\n";
    return 0;
}