小白书训练-Artificial Intelligence?

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=478

题意：实在是太挫了，这个题目做了好久。TAT，题意很简单，就是用语言描述了电压U电流I功率P中的任意两个，求另一个。

字符串转化以及单位判断嘛，不难，一直在WA，只是因为头文件用的是cstdio，用了stdio.h就恢复正常了，真是的，简直了！！！！

代码：

#include 
#include 

using namespace std;

char s[1000];

char *getnum(char *p,double &ans)
{
    ans = 0;
    while(*p >= '0' && *p <= '9')
    {
        ans = *p - '0' + ans * 10;
        p++;
    }

    if(*p == '.')
    {
        p++;

        double quan = 0.1;
        while(*p >= '0' && *p <= '9')
        {
            ans = ans + quan * (*p - '0');
            quan *= 0.1;
            p++;
        }
    }

    return p;
}

int main()
{
    int N;

    cin >> N;
    cin.get();

    double u,i,ap;
    bool bu,bi,bp;
    int no = 1;
    while(N--)
    {
        cin.getline(s,1000);
        char *p = s;

        bu = 0;
        bi = 0;
        bp = 0;
        while(*p != '\0')
        {
            if(*p == '=')
            {
                if(*(p - 1) == 'u' || *(p - 1) == 'U')
                {
                    bu = 1;
                    p++;
                    if(*p >= '0' && *p <= '9')
                        p = getnum(p,u);

                    if(*p == 'm')
                        u *= 0.001;
                    if(*p == 'k')
                        u *= 1000;
                    if(*p == 'M')
                        u *= 1000000;
                }else if(*(p - 1) == 'i' || *(p - 1) == 'I')
                {
                    bi = 1;
                    p++;
                    if(*p >= '0' && *p <= '9')
                        p = getnum(p,i);

                    if(*p == 'm')
                        i *= 0.001;
                    if(*p == 'k')
                        i *= 1000;
                    if(*p == 'M')
                        i *= 1000000;
                }else if(*(p - 1) == 'p' || *(p - 1) == 'P')
                {
                    bp = 1;
                    p++;
                    if(*p >= '0' && *p <= '9')
                        p = getnum(p,ap);

                    if(*p == 'm')
                        ap *= 0.001;
                    if(*p == 'k')
                        ap *= 1000;
                    if(*p == 'M')
                        ap *= 1000000;
                }
            }
            p++;
        }

        cout << "Problem #" << no++ << endl;

        if(bu && bi)
            printf("P=%.2lfW\n",u * i);
        else if(bp && bi)
            printf("U=%.2lfV\n",ap / i);
        else if(bu && bp)
            printf("I=%.2lfA\n",ap / u);
        cout << endl;
    }
    return 0;
}

梦续代码： http://www.hypo.xyz