分子量（Molar Mass, ACM/ICPC Seoul 2007, UVa1586）

给出一种物质的分子式（不带括号），求分子量。本题中分子式只包含4种原子，分别为C，H，O，N，原子量分别为12.01, 1.008, 16.00, 14.01（单位：g/mol）。例如，C6H5OH的分子量为94.108 (g/mol)。

本题要注意区分原子后不带数量（缺省数量为1）和带数量两种情况。

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

#define maxn 85
char buf[maxn];    // 输入的分子式字符串缓冲区

int find_index(const char* buffer, char ch) {
    for(int i = 0; buffer[i]; i++) 
        if(buffer[i] == ch) return i;
    return -1;
}

int main() {
    int T;
    const char * atom = "CHON";
    const float mass[] = { 12.01, 1.008, 16.00, 14.01 }; // C/H/O/N的原子量
    
    scanf("%d", &T);
    while (T--) {
        scanf("%s", buf);
        float tot = 0.0, unit = 0.0;
        int count = 0, i;
        bool q = false;
        for(i = 0; buf[i]; i++) {
            char ch = buf[i];
            if(isalpha(ch)) {
                tot += unit * count; // 结算上一个元素的质量
                int index = find_index(atom, toupper(ch));
                if(index != -1) {
                    unit = mass[index];
                    count = 1; // 原子数量缺省为1
                    q = true;
                } else { // 原子符号不在C/H/O/N之中
                    printf("WRONG ATOM INPUT!!\n");
                    break;
                }
            } else if(isdigit(ch)) {
                if(q) { // 此时原子数量缺省为1，改为计算实际的数量
                    count = 0;
                    q = false;
                }
                count = count * 10 + buf[i] - '0';
            }
        }
        if(!buf[i]) {
            tot += unit * count;
            printf("%.3f\n", tot);
        }
    }
    return 0;
}

原题说明如下

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