537 Artificial Intelligence?

解题思路都在程序注释里面,再次由于输出格式困扰很久,没法唉,题都是英文的,看着头疼,又只是看了大意.

关于最后一点注释 很重要的一点,网上搜索学习到的:double类sscanf用%lf,printf时用%f.

我参考别人代码时,发现很多人都用的%f来输出double,我用code::blocks gcc编译出来确实是不行,估计是编译器特性吧= =

差点忘记了: m (milli) = 10^-3, k (kilo) = 10^3 and M (Mega) = 10^6 

//感觉这道题麻烦之处在于,必须要知道读入的数据是哪一个物理量,还要转化单位,我没有相出什么捷径,所以用的笨办法,列出所有情况
//仔细做起来才发现,这题如果一个字符一个字符的读完一行处理,会比较纠结,在于究竟是读到哪里停止?\n?
//又想着用fgets读完一行,又想起这样读取那个数据很麻烦.
//最后思路是直接从头开读,每读到两组数据就处理,读完n组就结束
//又遇到困难了,用scanf是无法读入数据的,思考再三决定用数组读入那几个数,再用sscanf读.
//很重要的一点,网上搜索学习到的:double类sscanf用%lf,printf时用%f
#include<stdio.h>
#include<ctype.h>
#include<string.h>
int maxn = 100;
int main()
{
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
#endif
    int n, k = 1, num = 0;
    double U, I, P;
    int p = 0, u = 0, i = 0;//判断输入数据
    char c, s_num[maxn];
    scanf("%d\n", &n);
    while(k <= n)
    {
        c = getchar();
        if(c == 'P')
            if(getchar() == '=')
            {
                p = 1; num++;
                int j = 0;
                while(!isalpha(c = getchar())) s_num[j++] = c;
                sscanf(s_num, "%lf", &P);
                memset(s_num, '\0', maxn);
                if(c == 'm') P *= 0.001;
                if(c == 'k') P *= 1000;
                if(c == 'M') P *= 1000000;
            }
        if(c == 'U')
            if(getchar() == '=')
            {
                u = 1; num++;
                int j = 0;
                while(!isalpha(c = getchar())) s_num[j++] = c;
                sscanf(s_num, "%lf", &U);
                memset(s_num, '\0', maxn);
                if(c == 'm') U *= 0.001;
                if(c == 'k') U *= 1000;
                if(c == 'M') U *= 1000000;
            }
        if(c == 'I')
            if(getchar() == '=')
            {
                i = 1; num++;
                int j = 0;
                while(!isalpha(c = getchar())) s_num[j++] = c;
                sscanf(s_num, "%lf", &I);
                memset(s_num, '\0', maxn);
                if(c == 'm') I *= 0.001;
                if(c == 'k') I *= 1000;
                if(c == 'M') I *= 1000000;
            }

        if(num == 2) //输出
        {
            printf("Problem #%d\n", k);
            if(u == 0) printf("U=%.2fV", P/I);
            if(i == 0) printf("I=%.2fA", P/U);
            if(p == 0) printf("P=%.2fW", U*I);
            printf("\n\n");
            p = 0, u = 0, i = 0, num = 0;
            k++;
        }
    }
    return 0;
}