UVa1586 Molar Mass

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 (2n99) .

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

#include <iostream>
#include <string.h>
using namespace std;

#define maxn 100

float molar(char c)
{
	if(c == 'C')
	  return 12.01;
	else if(c == 'H')
	  return 1.008;
	else if(c == 'O')
	  return 16.00;
	else if(c == 'N')
	  return 14.01;
}

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
    	char s[maxn];
    	float total = 0;
    	scanf("%s", s);
    	
    	for(int i = 0; i < strlen(s); i++)
    	{
    		if(s[i] == 'C' || s[i] == 'H' || s[i] == 'O' || s[i] == 'N')
    		  total += molar(s[i]);
    		else
    		{
    			if((s[i] <= '9' && s[i] >= '1') && (s[i+1] <='0' || s[i+1] > '9'))
				  total += molar(s[i-1]) * (s[i]-'0'-1);//注意ASCII码中数字与正常数字，应减去'0'
				else
				{
					total += molar(s[i-1]) * ((s[i]-'0')*10 + (s[i+1]-'0'-1));
					//cout << ((s[i]-'0')*10 + (s[i+1]-'0'-1)) << endl;
					i++;
				}
			}
    		   
    		
		}
		printf("%.3f\n", total);
	}
	return 0;	
}