UVA-537 Artificial Intelligence?

题目大意：从一句话中抽取出已知条件，应用公式 P = U * I，求出未知的元素。

Sample Input

3
If the voltage is U=200V and the current is I=4.5A, which power is generated?
A light-bulb yields P=100W and the voltage is U=220V. Compute the current, please.
bla bla bla lightning strike I=2A bla bla bla P=2.5MW bla bla voltage?

Sample Output

Problem #1
P=900.00W
Problem #2
I=0.45A
Problem #3
U=1250000.00V

解题思路：对字符逐个判断，遇到 p,u,i 时，判断下一个是否是 = ，如果是，以 lf 读取已知条件的数值，运算后输出结果。

//UVA-537 Artificial Intelligence?#include<iostream>#include<cstdio>#include<cstring>using namespace std;char c;double u,i,p;int count = 1;char str[100000];int main() {    int T;    scanf("%d",&T);    getchar();    while (T--) {        u = 0, i = 0, p = 0;        int t = 2;        while (t--) {            while ((c = getchar()) != '\n' ) {                if (c == 'I') {                    if((c = getchar()) == '=') {                        scanf("%lf",&i);                        c = getchar();                        if(c == 'm')                            i = i * 0.001;                        if(c == 'k')                            i = i * 1000;                        if(c == 'M')                            i = i * 1000000;                        break;                    }                }                if(c == 'P') {                    if((c = getchar()) != '\n') {                        scanf("%lf",&p);                        c = getchar();                        if(c == 'm')                            p = p * 0.001;                        if(c == 'k')                            p = p * 1000;                        if(c == 'M')                            p = p * 1000000;                        break;                    }                }                if(c == 'U') {                    if((c = getchar()) != '\n') {                        scanf("%lf",&u);                        c = getchar();                        if(c == 'm')                            u = u * 0.001;                        if(c == 'k')                            u = u * 1000;                        if(c == 'M')                            u = u * 1000000;                        break;                    }                }            }        }        gets(str);        printf("Problem #%d\n",count++);        if (u == 0)            printf("U=%.2lfV",p/i);        if (i == 0)            printf("I=%.2lfA",p/u);        if (p == 0)            printf("P=%.2lfW",u*i);        printf("\n\n");    }    return 0;}