这道题一看感觉挺难，因为描述文字长篇大论，仔细读一读其实就是求相对分子质量。字符串处理。不过先来想偷懒用正则表达式，但是想了半天没有什么太好的方法。于是用普通的方法AC了。如果有大神有正则表达式的好方法留言一下，我也好学习学习。

import java.util.*;import java.util.regex.Matcher;import java.util.regex.Pattern;public class Main1586 {	public static void main(String[] args) {		Scanner scan = new Scanner(System.in);		int n = scan.nextInt();		while(n-- > 0) {			String str = scan.next();			double mass = 0;			for(int i=0; i
    
     ='C' && ch<='O') {					if(ch == 'C') {						int cnt = 0;						int flag = 0;						for(int j=i+1; j
     
      ='0' && str.charAt(j)<='9') {								cnt = cnt * 10 + (int)(str.charAt(j) - '0');								flag ++;							}							else break;						}						if(flag != 0)						    mass += (double)cnt * 12.01;						else							mass += 12.01;					}					if(ch == 'H') {						int cnt = 0;						int flag = 0;						for(int j=i+1; j
      
       ='0' && str.charAt(j)<='9') {								cnt = cnt * 10 + (int)(str.charAt(j) - '0');								flag ++;							}							else break;						}						if(flag != 0)						    mass += (double)cnt * 1.008;						else							mass += 1.008;					}					if(ch == 'O') {						int cnt = 0;						int flag = 0;						for(int j=i+1; j
       
        ='0' && str.charAt(j)<='9') {								cnt = cnt * 10 + (int)(str.charAt(j) - '0');								flag ++;							}							else break;						}						if(flag != 0)						    mass += (double)cnt * 16.00;						else							mass += 16.00;					}					if(ch == 'N') {						int cnt = 0;						int flag = 0;						for(int j=i+1; j
        
         ='0' && str.charAt(j)<='9') {								cnt = cnt * 10 + (int)(str.charAt(j) - '0');								flag ++;							}							else break;													}						if(flag != 0)						    mass += (double)cnt * 14.01;						else							mass += 14.01;					}				}			}			System.out.printf("%.3f\n", mass);				}	}}