[MM19-易] Surface to Volume Ratio(C++)

[MM19-易] Surface to Volume Ratio

Problem Description
Write a program to compute surface to volume ratio. We will be given the height, width, and depth of N cuboid, and determine the smallest surface to volume ratio among them.
Input File Format
The first line of the input data consists of N, where 0 < N <= 1000. The next N lines contain the height, width, and depth of each cuboid. All the dimensions are between 1 and 50.
Output Format
You should output the smallest surface to volume ratio as "a/b". This number must be simplified, that is, a and b must be prime to each other.
Example

Sample Input:	Sample Output:
2 2 3 4 1 1 1	13/6

1. #include <iostream>  
2. #include <string>  
3. #include <sstream>  
4. #include <stdio.h>  
5. #include <ctype.h>  
6. #include <cmath>  
7. using namespace std;  
8.   
9. int main() {  
10.     // [MM19-易] Surface to Volume Ratio  
11.     int n;  
12.     int a, b, c;  
13.     int mvol = 0, msur = 0;  
14.     cin >> n;  
15.     double min = 125000;  
16.     for(int i = 0;i < n;i++)  
17.     {  
18.         cin >> a >> b >> c;  
19.         int vol = a*b*c;  
20.         int sur = (a*b+b*c+a*c)*2;  
21.         double re = (double)sur/(double)vol;  
22.         if(re < min)  
23.         {  
24.             min = re;  
25.             mvol = vol;  
26.             msur = sur;  
27.         }  
28.     }  
29.       
30.     for(int i = 1;i < mvol;i++)  
31.     {  
32.         if(mvol % i == 0 && msur % i == 0)  
33.         {  
34.             if(i != 1)  
35.             {  
36.                 while(mvol % i == 0 && msur % i == 0)  
37.                 {  
38.                     mvol = mvol / i;  
39.                     msur = msur / i;  
40.                 }  
41.             }  
42.         }  
43.     }  
44.     cout << msur << "/" << mvol << endl;  
45.     return 0;  
46. }