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
[email protected]
Source
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;
}