Molar Mass, ACM/ICPC Seoul 2007,Uva1586 给出一种物质的分子式，求分子量

#include <stdio.h>
#include <string.h>
#define maxn 100
int main()
{
    char s[maxn];
    int i,j,v=0;
    memset(s, '0', sizeof(s));
    while(~scanf("%s", s))
    {
        double sum = 0.0;
        for( i = 0; i < strlen(s); i++)
        {
            v=0;
            switch(s[i])///在里面继续判断 如C10H10 C2H4
            {
            case 'C':
            {
                if(s[i+1] > '0' && s[i+1] <= '9')
                {
                    for(j=i+1; j<strlen(s); j++)
                    {
                        if(s[j]>='0'&&s[j]<='9')
                            v=10*v+(s[j]-48);
                        else
                            break;
                    }
                    sum += v* 12.01;
                }
                else
                    sum += 12.01;
                break;
            }
            case 'H':
            {
                if(s[i+1] > '0' && s[i+1] <= '9')
                {
                    for(j=i+1; j<strlen(s); j++)
                    {
                        if(s[j]>='0'&&s[j]<='9')
                            v=10*v+(s[j]-48);
                        else
                            break;
                    }
                    sum += v* 1.008;
                }
                else
                    sum += 1.008;
                break;
            }
            case 'O':
            {
                if(s[i+1] > '0' && s[i+1] <= '9')
                {
                    for(j=i+1; j<strlen(s); j++)
                    {
                        if(s[j]>='0'&&s[j]<='9')
                            v=10*v+(s[j]-48);
                        else
                            break;
                    }
                    sum += v* 16.00;
                }
                else
                    sum += 16.00;
                break;
            }
            case 'N':
            {
                if(s[i+1] > '0' && s[i+1] <= '9')
                {
                    for(j=i+1; j<strlen(s); j++)
                    {
                        if(s[j]>='0'&&s[j]<='9')
                            v=10*v+(s[j]-48);
                        else
                            break;
                    }
                    sum += v* 14.01;
                }
                else
                    sum += 14.01;
                break;
            }
            default:
                break;
            }
        }
        printf("%.3fg/mol\n", sum);
    }

    return 0;
}