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);}}
}