UVA - 1586 Molar mass

An organic compound is any member of a large class of chem-
ical 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 for-
mula, Dr. CHON wants to nd its molar mass. A molecular
formula, such as C3H4O3, identi es 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 ele-
ment, 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 rst 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

给出一种物质的分子式（不带括号），求分子量，只包含CHON四种原子。但是原子数为1的时候省略。而且分子数n最大可为99，所以需要判断字母后面两个字符。

[cpp]  view plain  copy

1. #include<iostream>   
2. #include<cstdio>  
3. #include<cstring>  
4. #include<ctype.h>  
5. #define maxn 100  
6. using namespace std;  
7. double switchit(char x)//将字母转换成相对应的原子量  
8. {  
9.     if(x=='C') return 12.01;  
10.     if(x=='H') return 1.008;  
11.     if(x=='O') return 16.00;  
12.     if(x=='N') return 14.01;  
13. }  
14. int main(){  
15. string s;  
16. int T;  
17. cin>>T;  
18. double ans;  
19. while(T--)  
20. {  
21.     cin>>s;  
22.     ans=0;  
23. for(int i=0;i<s.length();i++)  
24. {  
25.     if(s[i+1])  
26.     {if(isalpha(s[i])) { //当前位为字母  
27.         if(isalpha(s[i+1]))  ans+=switchit(s[i]);//后一位也为字母，直接加上当前位字母的原子量  
28. else if(isalpha(s[i+2])||!s[i+2]) {ans+=(switchit(s[i])*(s[i+1]-'0')); i++;}//后一位为数字，后两位为字母或者不存在，乘以后一位的数值  
29. else {  ans+=switchit(s[i])*((s[i+1]-'0')*10+s[i+2]-'0'); i+=2;   }//后面两个数都是数字，转化以后乘以原子量  
30.     }  
31.     }  
32.     else {ans+=switchit(s[i]);}//如果后一位不存在，直接加上当前位字母的原子量  
33. }  
34.   printf("%.3f\n",ans);  
35. }  
36.   return 0;  
37. }