UVA1586 Molar mass

DescriptionAn organic compound is any member of a large class of chemical compounds whose molecules contain carbon. Themolar 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 asC3H4O3 , identifies each constituent element by its chemical symbol and indicates the number of atoms of each element foundin 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 ofT 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 acapital letter and the length of the string is greater than 0 and less than 80. The quantity numbern 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

这道题最难得地方应该就是求出每个原子的个数，由题意可知每个原子个数不会超过99，所以只会是一位数或者两位数，所以遍历字符串每个字符，如果该字符后面是数字，则该原子个数就是后面的数字，若该字符后面不是数字，那么该原子个数就为1。

ac代码如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>int main(){int t,i,n;double sum;char s[85];scanf("%d",&t);while(t--){scanf("%s",s);sum=0;for(i=0;i<strlen(s);i++){if(s[i+1]>='0'&&s[i+1]<='9'){if(s[i+2]>='0'&&s[i+2]<='9'){n=(s[i+1]-'0')*10+(s[i+2]-'0');}elsen=s[i+1]-'0';}elsen=1;if(s[i]=='C')sum+=12.01*n;else if(s[i]=='H')sum+=1.008*n;else if(s[i]=='O')sum+=16.00*n;else if(s[i]=='N')sum+=14.01*n;}printf("%.3lf\n",sum);}return 0;}

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