hdu 3809

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Decrypt coordinate

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 703 Accepted Submission(s): 286

Problem Description

For national security, the coordinate systems of public map service are all in an encryption system. This means that when you got the latitude and longitude coordinates from a GPS device, you cannot finger out the real place at a public map with that coordinates.
We know an encryption system, which works as:
x1 = x – sqrt(y)
y1 = y – sqrt(x)
Now, we got some coordinates of POIs (point of interest, http://en.wikipedia.org/wiki/Point_of_interest, which you can just think as some points)
But even we have that formulations of how to encrypt coordinates, still we cannot decrypt any POI coordinate. Can you help us to decrypt all the POI coordinates?

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case consists of two float number(x, y) in a single line.

Technical Specification

1. 1 <= T <= 10000
2. 0 <= x, y <= 10000

Output

For each test case, output the case number first, then the decrypted coordinates with six decimal digits.
If multiply answer exists, output the one which has minimal x.

Sample Input

4 500.116 5000 0 0 10000 10000 5000 5000

Sample Output

Case 1: 570.995444 5023.895511 Case 2: 0.000000 0.000000 Case 3: 10100.501250 10100.501250 Case 4: 5071.212446 5071.212446

Author

momodi@WHU

Source

The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Preliminary

迭代法

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int t;
double x1,y11,x,y;
int main()
{
      int ca=1;
      scanf("%d",&t);
      while(t--)
      {
            scanf("%lf%lf",&x1,&y11);
            x=x1,y=y11;
            for(int i=0;i<45;i++)
            {
                  x=x1+sqrt(y);
                  y=y11+sqrt(x);
            }
            printf("Case %d: ",ca),ca++;
            printf("%.6lf %.6lf\n",x,y);
      }
      return 0;
}

hdu 3809

标签：des blog http ar io os sp for java

原文地址：http://www.cnblogs.com/a972290869/p/4166254.html

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