题目链接：
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;
}