uva 1586 - Molar mass

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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 NitrogenStandard 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 Input4CC6H5OHNH2CH2COOHC12H22O11Sample Output12.01094.10875.070342.296

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import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int n = in.nextInt();        String buffer = in.nextLine();        while ((n--) != 0) {            String s = in.next();            double count = 0;            int num = 0;            double sum = 0;            for (int i = 0; i < s.length(); i++) {                if (s.charAt(i) == 'C') {                    count = 12.01;                    num = 0;                } else if (s.charAt(i) == 'H') {                    count = 1.008;                    num = 0;                } else if (s.charAt(i) == 'O') {                    count = 16.00;                    num = 0;                } else if (s.charAt(i) == 'N') {                    count = 14.01;                    num = 0;                } else if( Integer.parseInt(s.charAt(i)+"") >= 0 && Integer.parseInt(s.charAt(i)+"") <= 9) {                    num++;                    if (num == 1) {                        sum += Integer.parseInt(s.charAt(i)+"") * count - 2*count;                    } else if (num == 2) {                        sum += ( Integer.parseInt(s.charAt(i)+"") + (Integer.parseInt(s.charAt(i-1)+"")*10)) * count - Integer.parseInt(s.charAt(i-1)+"")*count - count;                    }                }                sum += count;            }            System.out.printf("%.3f\n",sum);        }    }}

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小结

String s = in.next();

注意这行 和上个问题是一样的问题，如果是nextLine()， 程序会re，结果这样就好了，又是’\0’ 这样的问题，因为字符串默认会增加一个’\0’，所以又会判断，直接就报错了，re了很多次，内心是崩溃的，结果才发现这里错了。不断搜索原因也没发现。好了，终于ac了。

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