1586 - Molar mass

习题3-2 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。本题中的分子式只包含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94.108g/mol。

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

#define MAXN 85
#define C 12.01
#define H 1.008
#define O 16.00
#define N 14.01

int main(){
    int T=0;
    char s[MAXN];
    scanf("%d",&T);
    while(T--){
        double sum = 0.0;
        scanf("%s",s);
        int n = strlen(s);
        int i;
        for (i=0;i<n;){
            int cnt = 1;
            if (isdigit(s[i+1])){
                if (isdigit(s[i+2])){
                    cnt = (s[i+1]-'0')*10+(s[i+2]-'0');
                }
                else cnt = s[i+1]-'0';
            }
            switch (s[i]){
                case ('C'):
                    sum += C*cnt;
                    break;
                case ('O'):
                    sum += O*cnt;
                    break;  
                case ('N'):
                    sum += N*cnt;
                    break;
                case ('H'):
                    sum += H*cnt;
                    break;
                default:
                    break;
            }   
            i += (cnt==1?1:(cnt>9?3:2));
        }
        printf("%.3f\n",sum);
    }
    return 0;
}