题目1099题目信息运行结果本题排行讨论区
Lan Xiang’s Square
时间限制：1000ms | 内存限制：65535KB
难度：0
描述
Excavatortechnologywhichis strong,fastto Shandong to findLan Xiang.
Then the question comes.. 🙂
for this problem , i will give you four points. you just judge if they can form a square.
if they can, print "Yes", else print "No".
Easy ? just AC it.
输入T <= 105 cases.for every casefour points, and every point is a grid point .-10^8 <= all interger <= 10^8。grid point is both x and y are interger.输出Yes or No样例输入11 1-1 1-1 -11 -1样例输出Yes提示you think this is a easy problem ? you dare submit, i promise you get a WA. :)来源myself上传者
ACM_张开创
题意：给出四个点坐标，判断是否能组成正方形。
证明：四条边相等，并且有一个角是直角 为正方形。
题很简单 ，自己的思路也一直整错 可就是wa
可能数据太给力了，调试n次，发现sqrt()导致精度损失。 这个错误可真难找。。。
先贴上wa的代码
#include <stdio.h>#include <math.h>#define inf 500000005int main(){double a[4],b[4],x1,x2,x3,x4,y1,y2,y3,y4;double sum1,sum2,l1,l2,l3,l4,l;int i,t;scanf("%d",&t);while(t–){sum1=sum2=0,x1=x2=x3=x4=y1=y2=y3=y4=inf;for(i=0;i<4;i++)scanf("%lf %lf",&a[i],&b[i]),sum1+=a[i],sum2+=b[i];sum1=sum1/4,sum2=sum2/4;for(i=0;i<4;i++){if(a[i]<sum1&&b[i]>=sum2)x1=a[i],y1=b[i];if(a[i]<=sum1&&b[i]<sum2)x2=a[i],y2=b[i];if(a[i]>=sum1&&b[i]>sum2)x3=a[i],y3=b[i];if(a[i]>sum1&&b[i]<=sum2)x4=a[i],y4=b[i];}if(x1==inf||x2==inf||x3==inf||x4==inf||y1==inf||y2==inf||y3==inf||y4==inf){printf("No\n");continue;}l1=sqrt(pow(y1-y2,2)+pow(x1-x2,2));l2=sqrt(pow(y2-y4,2)+pow(x2-x4,2));l3=sqrt(pow(y3-y4,2)+pow(x3-x4,2));l4=sqrt(pow(y1-y3,2)+pow(x1-x3,2));l=sqrt(pow(y3-y2,2)+pow(x3-x2,2));if(l1==l2&&l1==l3&&l1==l4&&fabs(l1*l1+l4*l4-l*l)<0.00001)printf("Yes\n");elseprintf("No\n");}return 0;}ac代码
#include <stdio.h>#include <math.h>#define inf 500000005int main(){double a[4],b[4],x1,x2,x3,x4,y1,y2,y3,y4;double sum1,sum2,l1,l2,l3,l4,l;//l1,l2,l3,l4,l分别表示四条边和一个对角线。int i,t;scanf("%d",&t);while(t–){sum1=sum2=0,x1=x2=x3=x4=y1=y2=y3=y4=inf;//初始坐标for(i=0;i<4;i++)scanf("%lf %lf",&a[i],&b[i]),sum1+=a[i],sum2+=b[i];sum1=sum1/4,sum2=sum2/4;for(i=0;i<4;i++)//更新坐标{if(a[i]<sum1&&b[i]>=sum2)x1=a[i],y1=b[i];if(a[i]<=sum1&&b[i]<sum2)x2=a[i],y2=b[i];if(a[i]>=sum1&&b[i]>sum2)x3=a[i],y3=b[i];if(a[i]>sum1&&b[i]<=sum2)x4=a[i],y4=b[i];}if(x1==inf||x2==inf||x3==inf||x4==inf||y1==inf||y2==inf||y3==inf||y4==inf)//如果有一个坐标未更新 ，no{printf("No\n");continue;}l1=pow(y1-y2,2)+pow(x1-x2,2);//和wa不同的地方l2=sqrt(pow(y2-y4,2)+pow(x2-x4,2));l3=sqrt(pow(y3-y4,2)+pow(x3-x4,2));l4=pow(y1-y3,2)+pow(x1-x3,2);//同上l=pow(y3-y2,2)+pow(x3-x2,2);//同上if(sqrt(l1)==l2&&sqrt(l1)==l3&&sqrt(l1)==sqrt(l4)&&l1+l4==l)printf("Yes\n");elseprintf("No\n");}return 0;}
，经验是由痛苦中粹取出来的
