例题3-2

An 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 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 NameCarbonHydrogenOxygenNitrogenStandard Atomic Weight12.01 g/mol1.008 g/mol16.00 g/mol14.01 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 numbern 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

本题难度不大，主要在于情况的细分，其中最重要的是对#include<iomanip>头文件的应用，注意对fixed，setprecision(3)的应用

#include <iostream>#include<cstring>#include<iomanip>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */char s[81]; int main(int argc, char** argv) {int t;cin>>t;while(t>=0){t--;cin.getline(s,sizeof(s));int i=0;double sum=0;while(s[i]!='\0'){   double now=0;if(s[i]=='C')now=12.010;if(s[i]=='H')now=1.008;if(s[i]=='O')now=16.00;if(s[i]=='n')now=14.01;if(s[i+1]<='9'&&s[i+1]>='0'){now=now*(s[i+1]-'0');i+=2;}else i++;sum+=now;if(s[i]=='\0'){cout<<fixed<<setprecision(3)<<sum<<endl;}}}return 0;}