Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 291358 Accepted Submission(s): 100298
1 #include<math.h> 2 #include<stdio.h> 3 int main() 4 { 5 double x1,y1,x2,y2,m; 6 while(~scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)) 7 //等效于 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)!=EOF 可以少打几个字了··· 8 { 9 m=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 10 printf("%.2lf\n",m); 11 } 12 return 0; 13 }
原文:https://www.cnblogs.com/lightice/p/10261040.html