codeforce 851 B Arpa and an exam about geometry(思路)

B. Arpa and an exam about geometry
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input
The only line contains six integers ax, ay, bx, by, cx, cy (|ax|, |ay|, |bx|, |by|, |cx|, |cy| ≤ 109). It’s guaranteed that the points are distinct.

Output
Print “Yes” if the problem has a solution, “No” otherwise.

You can print each letter in any case (upper or lower).

Examples
input
0 1 1 1 1 0
output
Yes
input
1 1 0 0 1000 1000
output
No
Note
In the first sample test, rotate the page around (0.5, 0.5) by .

In the second sample test, you can’t find any solution.

题意：给出a,b,c三个点，问是否存在一点，以此点旋转相同的角度，a可以到达b，b可以到达c
分析得 ab=bc,且abc都不在一条线上满足条件
计算abc不在一条线，则计算 ab斜率!=bc斜率，考虑一下除0的情况

#include<bits/stdc++.h>using namespace std;typedef long long LL;LL solve(LL ax,LL ay,LL bx,LL by){    return (ax-bx)*(ax-bx)+(ay-by)*(ay-by);}bool check(LL ax,LL ay,LL bx,LL by,LL cx,LL cy){    return (by-ay)*(cx-bx)==(bx-ax)*(cy-by);}int main(){    LL ax,ay,bx,by,cx,cy;    scanf("%lld%lld%lld%lld%lld%lld",&ax,&ay,&bx,&by,&cx,&cy);    if(solve(ax,ay,bx,by)==solve(bx,by,cx,cy))    {        if(check(ax,ay,bx,by,cx,cy))            printf("No\n");        else            printf("Yes\n");    }    else    printf("No\n");}