(hdu 簡單題 128道)hdu 2001 計算兩點間的距離
阿新 • • 發佈:2019-01-25
題目:
計算兩點間的距離
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 115566 Accepted Submission(s): 44177
Problem Description輸入兩點座標(X1,Y1),(X2,Y2),計算並輸出兩點間的距離。
Input輸入資料有多組,每組佔一行,由4個實陣列成,分別表示x1,y1,x2,y2,資料之間用空格隔開。
Output對於每組輸入資料,輸出一行,結果保留兩位小數。
Sample Input0 0 0 1 0 1 1 0
Sample Output1.00 1.41
Authorlcy
Source
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題目分析:
簡單題。
程式碼如下:
/* * g.cpp * * Created on: 2015年3月20日 * Author: Administrator hdu 2001 */ #include <iostream> #include <cstdio> #include <cmath> struct PPoint{//結構體儘量不要定義成Point這種,容易和C/C++本身中的變數同名 double x; double y; PPoint(double _x = 0,double _y = 0):x(_x),y(_y){ } PPoint operator - (const PPoint& op2) const{ return PPoint(x - op2.x,y - op2.y); } double operator^(const PPoint &op2)const{ return x*op2.y - y*op2.x; } }; inline double sqr(const double &x){ return x*x; } inline double dis2(const PPoint &p0,const PPoint &p1){ return sqr(p0.x - p1.x) + sqr(p0.y - p1.y); } inline double dis(const PPoint& p0,const PPoint& p1){ return sqrt(dis2(p0,p1)); } int main(){ PPoint p1,p2; while(scanf("%lf %lf %lf %lf",&p1.x,&p1.y,&p2.x,&p2.y)!=EOF){ printf("%.2lf\n",dis(p1,p2)); } return 0; }