IEEEXtreme 10.0 - Counting Molecules

这是 meelo 原创的 IEEEXtreme极限编程大赛题解

Xtreme 10.0 - Counting Molecules

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。用矩阵的形式可以表示成下面的样子：

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。

判断一个分数是否是一个整数，有以下两种办法

• 判断 分母 % 分子 == 0
• 用浮点数的除法，然后取整，判断是否相等

程序

C++

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    long long c, h, o;
    cin >> c >> h >> o;
    if((-*c+h+*o)>= && (-*c+h+*o)%== &&
      (-h+*o)>= && (-h+*o)%== &&
      (*c+h-*o)>= && (*c+h-*o)%==) {
        printf("%lld %lld %lld", (-*c+h+*o)/, (-h+*o)/, (*c+h-*o)/);
    }
    else {
        printf("Error");
    }
    return ;
}

Python2

x, y, z = map(int, raw_input().split())
a = ((2 * z) - (4 * x) + y) / 4.0
b = ((2 * z) - y) / 4.0
c = (x - b) / 6.0

if a != a // 1 or a < 0:
    print "Error"
elif b != b // 1 or b < 0:
    print "Error"
elif c != c // 1 or c < 0:
    print "Error"
else:
    print int(round(a)),int(round(b)),int(round(c))

from: hackerranksolutionsforprogrammers.blogspot.com/2016/10/counting-molecules-by-ieeextreme.html

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