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