分子量（Molar Mass UVa1586）

题目来自张汝佳的《算法竞赛入门经典（第二版）》

题目描述：

我的代码：

#include<iostream>
#include<cstring>
#include <iomanip>
using namespace std;
#define C 12.01
#define H 1.008
#define O 16.00
#define N 14.01
int main()
{
    int i, k;
    char enter[1000];
    cin >> k;
    while (k--)
    {
        cin >> enter;
        double tot = 0.0;
        int num = 0, j;
        double ans = 0;
        for (i = 0; i < strlen(enter); i++)
        {
            ans = 0;
            switch (enter[i])
            {
            case ‘C‘:
                num = 0;
                if (enter[i + 1] <= ‘9‘ && enter[i + 1] >= ‘0‘)
                {
                    i++;
                    while (enter[i] <= ‘9‘ && enter[i] >= ‘0‘)
                    {
                        if (num == 0)
                            num = i;
                        i++;
                    }

                    for (j = i; num < j; num++)
                        ans += ((double)(enter[num] - ‘0‘) * pow(10, j - num - 1));

                    tot += ans * C;
                    i--;
                }
                else
                    tot += C;
                break;
            case ‘H‘:
                num = 0;
                if (enter[i + 1] <= ‘9‘ && enter[i + 1] >= ‘0‘)
                {
                    i++;
                    while (enter[i] <= ‘9‘ && enter[i] >= ‘0‘)
                    {
                        if (num == 0)
                            num = i;
                        i++;
                    }

                    for (j = i; num < j; num++)
                        ans += ((double)(enter[num] - ‘0‘) * pow(10, j - num - 1));

                    tot += ans * H;
                    i--;
                }
                else
                    tot += H;
                break;
            case ‘O‘:
                num = 0;
                if (enter[i + 1] <= ‘9‘ && enter[i + 1] >= ‘0‘)
                {
                    i++;
                    while (enter[i] <= ‘9‘ && enter[i] >= ‘0‘)
                    {
                        if (num == 0)
                            num = i;
                        i++;
                    }

                    for (j = i; num < j; num++)
                        ans += ((double)(enter[num] - ‘0‘) * pow(10, j - num - 1));

                    tot += ans * O;
                    i--;
                }
                else
                    tot += O;
                break;
            case ‘N‘:
                num = 0;
                if (enter[i + 1] <= ‘9‘ && enter[i + 1] >= ‘0‘)
                {
                    i++;
                    while (enter[i] <= ‘9‘ && enter[i] >= ‘0‘)
                    {
                        if (num == 0)
                            num = i;
                        i++;
                    }

                    for (j = i; num < j; num++)
                        ans += ((double)(enter[num] - ‘0‘) * pow(10, j - num - 1));

                    tot += ans * N;
                    i--;
                }
                else
                    tot += N;
                break;
            }
        }
        cout << setiosflags(ios::fixed) << setprecision(3) << tot << "g/mol" << endl;
    }
    return 0;
}

Sample Input

4 
C 
C6H5OH 
NH2CH2COOH 
C12H22O11

Sample Output

12.010 
94.108 
75.070 
342.296

水题，但是有陷阱，所以特地放了上来╮(╯▽╰)╭，求大神们更好的代码~