Molar mass UVA - 1586

​ 　 An organic compound is any member of a large class of chemical compounds whose molecules contain carbon. The molar mass of an organic compound is the mass of one mole of the organic compound. The molar mass of an organic compound can be computed from the standard atomic weights of the elements.

​ 　 When an organic compound is given as a molecular formula, Dr. CHON wants to find its molar mass. A molecular formula, such as C3H4O3, identifies each constituent element by its chemical symbol and indicates the number of atoms of each element found in each discrete molecule of that compound. If a molecule contains more than one atom of a particular element, this quantity is indicated using a subscript after the chemical symbol.

​ 　 In this problem, we assume that the molecular formula is represented by only four elements, ‘C’ (Carbon), ‘H’ (Hydrogen), ‘O’ (Oxygen), and ‘N’ (Nitrogen) without parentheses.

​ 　 The following table shows that the standard atomic weights for ‘C’, ‘H’, ‘O’, and ‘N’.

Atomic Name	Carbon	Hydrogen	Oxygen	Nitrogen
Standard Atomic Weight	12.01 g/mol	1.008 g/mol	16.00 g/mol	14.01 g/mol

​ 　 For example, the molar mass of a molecular formula C6H5OH is 94.108 g/mol which is computed by 6 × (12.01 g/mol) + 6 × (1.008 g/mol) + 1 × (16.00 g/mol).

​ 　 Given a molecular formula, write a program to compute the molar mass of the formula.

Input

​ 　 Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case is given in a single line, which contains a molecular formula as a string. The chemical symbol is given by a capital letter and the length of the string is greater than 0 and less than 80. The quantity number n which is represented after the chemical symbol would be omitted when the number is 1 (2 ≤ n ≤ 99).

Output

​ 　 Your program is to write to standard output. Print exactly one line for each test case. The line should contain the molar mass of the given molecular formula.

Sample Input

4
C
C6H5OH
NH2CH2COOH
C12H22O11

Sample Output

12.010
94.108
75.070
342.296

HINT

​ 　 题目大体意思是让你求一个有机分子的摩尔质量，问题的关键是如何求得各个有机分子得各个原子得总数。这里由于每一个原子的总数小于100个，因此可以采用直接判断的方法来求出总数。当然也可以使用递归函数来求出总数，但那就有点不划算了。

Accepted

#include<stdio.h>
#include<string.h>

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		char molar[81];
		int  num[4] = { 0 };					//num数组是用来保存各个原子的总数，下标为0 1 2 3的元素分别是C H O N的总数
		scanf("%s", molar);
		int len = strlen(molar);
		for (int i = 0;i < len;i++)
		{
			if (molar[i] <= '9' && molar[i] >= '0')continue;
			int k;
			if (molar[i] == 'C')k = 0;
			else if (molar[i] == 'H')k = 1;
			else if (molar[i] == 'O')k = 2;
			else if (molar[i] == 'N')k = 3;

			num[k]++;							//首先加上一是为了防止元素后面就是元素的情况，如果后面是数字就再后面减去一抵消
			if (i + 1 < len && molar[i + 1] <= '9' && molar[i + 1] >= '0')
			{
				if (i + 2 < len && molar[i + 2] <= '9' && molar[i + 2] >= '0')
					num[k] += (molar[i + 1] - '0') * 10 + molar[i + 2] - '0' - 1;
				else num[k] += molar[i + 1] - '0' - 1;
			}
		}
		double weigh = num[0] * 12.01 + num[1] * 1.008 + num[2] * 16.00 + num[3] * 14.01;
		printf("%.3lf\n", weigh);				//求出总质量
	}

}