技術標籤：C語言c++c

Mad scientist Mike is busy carrying out experiments in chemistry. Today he will attempt to join three atoms into one molecule.

A molecule consists of atoms, with some pairs of atoms connected by atomic bonds. Each atom has a valence number — the number of bonds the atom must form with other atoms. An atom can form one or multiple bonds with any other atom, but it cannot form a bond with itself. The number of bonds of an atom in the molecule must be equal to its valence number.

Mike knows valence numbers of the three atoms. Find a molecule that can be built from these atoms according to the stated rules, or determine that it is impossible.

Input
The single line of the input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 106) — the valence numbers of the given atoms.

Output
If such a molecule can be built, print three space-separated integers — the number of bonds between the 1-st and the 2-nd, the 2-nd and the 3-rd, the 3-rd and the 1-st atoms, correspondingly. If there are multiple solutions, output any of them. If there is no solution, print “Impossible” (without the quotes).

Examples
Input
1 1 2
Output
0 1 1
Input
3 4 5
Output
1 3 2
Input
4 1 1
Output
Impossible
Note
The first sample corresponds to the first figure. There are no bonds between atoms 1 and 2 in this case.

The second sample corresponds to the second figure. There is one or more bonds between each pair of atoms.

The third sample corresponds to the third figure. There is no solution, because an atom cannot form bonds with itself.

The configuration in the fourth figure is impossible as each atom must have at least one atomic bond.

思路：首先滿足的條件：
1.所有原子的階數之和必須時偶數，否則就會有單鍵多出來；
2.其次，自己不能和自己相接，所以，不能出現一個原子的階數大於另外兩個階數之和。如果大於，多出來的階數只能和自己相接，不成立；
核心：
1.首先，要明白，鍵數與原子階數的關係，後者為前者的兩倍；不妨設階數為a，b，c，a和b之間鍵數為x；b和c為y；a和c為z；就有x+y+z==(a+b+c)/2…(1)
下面的方程（這就不要解釋怎麼來的了吧）
a=x+z;…(2)
b=x+y;…(3)
c=z+y;…(4)
解上面列的(1)(2)(3)(4)方程得：
x=(a+b-c)/2;
y=(b+c-a)/2;
z=(a+c-b)/2;
ac程式碼：

#include<bits/stdc++.h>
#define LL long long
using namespace std;
int main()
{
	LL a,b,c,x,y,z;
	cin>>a>>b>>c;
	x=(a+b-c)/2;
	y=(b+c-a)/2;
	z=(a+c-b)/2;
	if(x<0||y<0||z<0||(a+b+c)%2) cout<<"Impossible"<<endl;
	else cout<<x<<" "<<y<<" "<<z<<endl;
	return 0;
}

總結：一開始找不到關係，也就沒有思路，這tm就是變相在考化學，所以學acm還要是個全能人才才行。。。。哈哈 : D