Xtreme 10.0 - Counting Molecules

这是 meelo 原创的 IEEEXtreme极限编程大赛题解

题目来源 第10届IEEE极限编程大赛
https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/counting-molecules

Your task is to count the number of molecules in a cup of soda which contains distilled water, carbon dioxide, andglucose. You have a machine that counts the number of atoms of carbon, hydrogen, and oxygen in a given sample.

Input Format

The input consists of a single line with three space separated integers: c, h, and o
where
c is the count of carbon atoms
h is the count of hydrogen atoms
o is the count of oxygen atoms

Constraints

0 ≤ c, h, o < 1010

Output Format

If the number of atoms is consistent with a mixture containing only water, carbon dioxide, and glucose molecules, the output should consist of a single line containing three space separated integers: the number of water molecules, the number of carbon dioxide molecules, and the number of glucose molecules.
If the number of atoms is not consistent with a mixture containing only water, carbon dioxide, and glucose molecules, the output should consist of a line containing the word Error

Sample Input

10 0 20

Sample Output

0 10 0

Explanation

The input indicates that there are 10 carbon atoms and 20 oxygen atoms. The only way that this could occur would be if there were 0 water molecules, 10 carbon dioxide molecules, and 0 glucose molecules.
Note that there are additional sample inputs available if you click on the Run Code
button.

题目解析
这题就是求解一个三元方程组。

三个未知数，三个方程。同时三个方程线性无关，有唯一解。

由于物质的个数是非负整数，约束方程组的解是非负整数。

程序
C++

#include 
#include 
#include 
#include 
#include 
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    long long c, h, o;
    cin >> c >> h >> o;
    if((-4*c+h+2*o)>=0 && (-4*c+h+2*o)%4==0 &&
      (-h+2*o)>=0 && (-h+2*o)%4==0 &&
      (4*c+h-2*o)>=0 && (4*c+h-2*o)%24==0) {
        printf("%lld %lld %lld", (-4*c+h+2*o)/4, (-h+2*o)/4, (4*c+h-2*o)/24);
    }
    else {
        printf("Error");
    }
    return 0;
}

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