UVa-1586 - Molar mass

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

代码很臃肿，看着不爽请无情批判

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

float r(int t, char c) {
float v = 0;
switch (c)
{
case 'C':v = (float)t*12.01;
break;
case 'H':v = (float)t*1.008;
break;
case 'N':v = (float)t*14.01;
break;
case 'O':v = (float)t*16.00;
break;
default:
break;
}
return v;
}
int main() {
int T;
char s[85];
scanf("%d",&T);
while(T--) {
float sum = 0;
int flag = 1;
memset(s,'a',sizeof(s));
scanf("%s",s);
int n = strlen(s), t = 0;
char c = s[0];
if (n == 1)sum += r(1, c);
else if (n > 1) {
for (int i = 1;i < n;i++) {
if (s[i] >= '0'&&s[i] <= '9') {
t = t * 10 + s[i] - '0';
flag = 0;
}
if (s[i] >= 'A'&&s[i] <= 'Z'||i==n-1) {
if (flag != 0)t = 1;
sum += r(t, c);
flag = 1;
t = 0;
c = s[i];
}
if (s[i] >= 'A'&&s[i] <= 'Z' && i == n - 1) {
sum += r(1,s[i]);
}
}
}
printf("%.3f\n",sum);
}
return 0;
}