输入
输入的第一行是一个正整数n,表示有n组测试数据。
接下来n行每行输入一个字符串,表示某个分子式,分子式中只包含大写字母和数字。
注意:
输入数据只包含8种元素,而这8种元素的相对原子质量如下:
H(1),C(12),N(14),O(16),F(19),P(31),S(32),K(39)。
输出
对于每组输入,输出相对分子质量。
样例输入
4
H2O
KOH
CH4
SO2
样例输出
18
56
16
64
代码
import java.util.Scanner;import java.util.regex.Pattern;public class Main2 {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();String[] strings1 = new String[n];for (int i = 0; i < n; i++)strings1[i] = sc.next();for (int i = 0; i < n; i++) {String[] strings2 = strings1[i].split("");int sum = 0;for (int j = 0; j < strings2.length; j++) {switch (strings2[j]) {case "H":if (j < strings2.length - 1 &&Pattern.matches("\\d", strings2[j + 1])) {sum += (Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 1;}break;case "C":if (j < strings2.length - 1 &&Pattern.matches("\\d", strings2[j + 1])) {sum += (12 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 12;}break;case "N":if (j < strings2.length - 1 &&Pattern.matches("\\d", strings2[j + 1])) {sum += (14 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 14;}break;case "O":if (j < strings2.length - 1 && Pattern.matches("\\d", strings2[j + 1])) {sum += (16 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 16;}break;case "F":if (j < strings2.length - 1 && Pattern.matches("\\d", strings2[j + 1])) {sum += (19 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 19;}break;case "P":if (j < strings2.length - 1 && Pattern.matches("\\d", strings2[j + 1])) {sum += (31 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 31;}break;case "S":if (j < strings2.length - 1 && Pattern.matches("\\d", strings2[j + 1])) {sum += (32 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 32;}break;case "K":if (j < strings2.length - 1 && Pattern.matches("\\d", strings2[j + 1])) {sum += (39 * Integer.parseInt(strings2[j + 1]));j += 1;} else {sum += 39;}break;default:break;}}System.out.println(sum);}}}
小明最近迷上了化学 几乎天天在实验室做实验 但是很多实验生成的化学产物的相对分子质量令他很困惑 不知如何计算 请你编程帮他计算