HDU2073 Infinite Road "water problem" "recursion"

The Infinite Road

Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) total submission (s): 6113 Accepted S Ubmission (s): 3097
Problem Description
Sweet since childhood like to draw pictures, recently he bought a smart brush, because just contact, so sweet will only use it to draw a straight line, so he in the plane Cartesian coordinate system to draw the following graphic:
Sweet friend Honey found the above figure is still a bit of a rule, so he asked the sweet: in your drawing, I give you two points, please calculate a connection two points of the length of the line (that is, the length of the route along the line) bar.

Input
The first number is a positive integer n (≤100). The number of groups representing the data.
Each set of data consists of four non-negative integers x1,y1,x2,y2; all numbers are not greater than 100.

Output
For each set of data, the line distance between two points (X1,y1) and (x2,y2) is output. Note that the output is accurate to 3 digits after the decimal point.

Sample Input
5
0 0 0 1
0 0 1 0
2 3 3 1
99 99 9 9
5 5 5 5

Sample Output
1.000
2.414
10.646
54985.047
0.000

Author
Lily

Source

Zhejiang University of Technology Network tryouts

The main topic: as shown in the figure, give you two coordinates, to find two points between the line distance.

Idea: Using recursion. Here, the process of recursion from (0,0) point to (x, y) point polyline is simulated.

Set I as horizontal axis and J as ordinate. The path is first right down to the vertical axis of 0, then one step to the horizontal

Coordinates are 0, and the ordinate is the position of the original horizontal axis +1.

so the direct simulation of recursive derivation. Use Ans[x][y] to save (0,0) The distance from the point to (x, y) points. Ultimately

The result is: Fabs (Ans[x1][y1]-ans[x2][y2]).

#include <iostream> #include <algorithm> #include <cstdio> #include <cmath>using namespace std; Double Ans[220][220];int Main () {for    (int j = 1; J <; J + +)    {for        (int i = 0; I <= J; i++)        {            if (i = = 0)                ANS[I][J] = Ans[j-1][i] + sqrt (1.0*j*j+1.0* (j-1) * (j-1));            else                ans[i][j-i] = ans[i-1][j-i+1] + sqrt (2.0);        }    }    int N;    int x1,y1,x2,y2;    Cin >> N;    while (n--)    {        cin >> x1 >> y1 >> x2 >> y2;        printf ("%.3lf\n", Fabs (Ans[x1][y1]-ans[x2][y2]));    }    return 0;}

HDU2073 Infinite Road "water problem" "recursion"