Codeforces Round #432 Div2 B. Arpa and an exam about geometry
2017-08-30 16:55
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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.
题目意思就是给出a,b,c三个点的坐标,然后绕着某个点旋转问是否能够a点变成b点,b点变成c点。若能输出Yes,否则输出No。
思路:判断三个点是否在一条直线上,若在一条直线上则不能。再判断a到b点的距离是否和b和c点的相等,若相等且不在一条直线上则可以。
40的数据:
0 0 1000000000 1 1000000000 -999999999
输出No.
需要注意的一点为需要用long long int 。一开始用的double精度不够过不了。
其他一些数据:
589824 196608 262144 196608 0 0
输出 Yes.
999999999 1000000000 0 0 -1000000000 -999999999
输出 Yes.
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.
题目意思就是给出a,b,c三个点的坐标,然后绕着某个点旋转问是否能够a点变成b点,b点变成c点。若能输出Yes,否则输出No。
思路:判断三个点是否在一条直线上,若在一条直线上则不能。再判断a到b点的距离是否和b和c点的相等,若相等且不在一条直线上则可以。
40的数据:
0 0 1000000000 1 1000000000 -999999999
输出No.
需要注意的一点为需要用long long int 。一开始用的double精度不够过不了。
其他一些数据:
589824 196608 262144 196608 0 0
输出 Yes.
999999999 1000000000 0 0 -1000000000 -999999999
输出 Yes.
#include<stdio.h> #include<string.h> #include<algorithm> #include<cmath> #include<cctype> #include<iostream> using namespace std; int main() { long long int ax,ay,bx,by,cx,cy; scanf("%lld%lld%lld%lld%lld%lld",&ax,&ay,&bx,&by,&cx,&cy); long long int l1=((by-ay)*(by-ay)+(bx-ax)*(bx-ax)); long long int l2=((cy-by)*(cy-by)+(cx-bx)*(cx-bx)); int mark=0; // printf("l1=%lf,l2=%lf\n",l1,l2); // printf("by-ay=%lf\n",by-ay); if((ax==bx&&ax==cx)||(ay==by&&ay==cy)) mark=1; else { double k1=(by-ay)*1.0/(bx-ax); double k2=(cy-by)*1.0/(cx-bx); if(k1==k2) mark=1; } // printf("%d\n",mark); if(!mark&&l1==l2) printf("Yes\n"); else printf("No\n"); return 0; }
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