HDOJ 2001 计算两点间的距离

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计算两点间的距离

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 94488 Accepted Submission(s): 36264

Problem Description

输入两点坐标（X1,Y1）,（X2,Y2）,计算并输出两点间的距离。

Input

输入数据有多组，每组占一行，由4个实数组成，分别表示x1,y1,x2,y2,数据之间用空格隔开。

Output

对于每组输入数据，输出一行，结果保留两位小数。

Sample Input

0 0 0 10 1 1 0

Sample Output

1.001.41

Author

lcy

第一次代码：

#include <stdio.h>#include <math.h>int main (){    double x1,x2,y1,y2,d;    while (scanf("%lf%lf%lf%lf",&x1,&x2,&y1,&y2)!=EOF)    {        d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));        printf ("%.2lf\n",d);    }    return 0;}

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor111057402014-07-20 17:49:31Wrong Answer20010MS240K251 BC++Hidden Secret

第二次代码：

#include <stdio.h>#include <math.h>int main (){    double x1,x2,y1,y2;    while (scanf("%lf%lf%lf%lf",&x1,&x2,&y1,&y2)!=EOF)    {        printf ("%.2lf\n",sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));    }    return 0;}

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor111057902014-07-20 17:55:37Wrong Answer200115MS240K235 BC++Hidden Secret

第三次代码：

#include <cstdio>#include <cmath>int main(){    double x1, y1, x2, y2;    while(scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2) != EOF)        printf("%.2lf\n", sqrt((double)((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1))));    return 0;}

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor111058182014-07-20 17:59:19Accepted20010MS240K239 BC++Hidden Secret

第四次代码：

#include <cstdio>#include <cmath>int main(){    double x1, y1, x2, y2;    while(scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2) != EOF)        printf("%.2lf\n", sqrt((double)((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1))));    return 0;}

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor111058282014-07-20 18:00:28Accepted20010MS240K236 BC++Hidden Secret

第五次代码：

#include <stdio.h>#include <math.h>int main (){    double x1,x2,y1,y2,d;    while (scanf("%lf%lf%lf%lf",&x1,&x2,&y1,&y2)!=EOF)    {        d=(double)sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));        printf ("%.2lf\n",d);    }    return 0;}

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor111058722014-07-20 18:05:16Wrong Answer200115MS240K259 BC++Hidden Secret

表示此题始终不知道自己错在哪里

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