Problem A: Arti cial Intelligence? 结题报告

Problem A: Arti cial Intelligence?


Description

Physics teachers in high school often think that problems given as text are more demanding thanpure computations. After all, the pupils have to read and understand the problem rst! So they don't state a problem like U=10V, I=5A, P=?" but rather like You have an electrical circuit that contains a battery with a voltage of U=10V and a light-bulb. There's an electrical current of I=5A through the bulb. Which power is generated in the bulb?". However, half of the pupils just don't pay attention to the text anyway. They just extract from the text what is given: U=10V, I=5A. Then they think: Which formulae do I know? Ah yes, P=U*I. Therefore P=10V*5A=500W. Finished." OK, this doesn't always work, so these pupils are usually not the top scorers in physics tests. But at least this simple algorithm is usually good enough to pass the class. (Sad but true.) Today we will check if a computer can pass a high school physics test. We will concentrate on the P-U-I type problems rst. That means, problems in which two of power, voltage and current are given

and the third is wanted. Your job is to write a program that reads such a text problem and solves it according to the simple algorithm given above.

 

Input

The rst line of the input le will contain the number of test cases. Each test case will consist of one line containing exactly two data elds and some additional arbitrary words. A data eld will be of the form I=xA, U=xV or P=xW, where x is a real number. Directly before the unit (A, V or W) one of the pre xes m (milli), k (kilo) and M (Mega) may also occur. To summarize it: Data elds adhere to the following grammar:

DataFiled ::= Concept '=' RealNumber [Profix] Unit

Concept   ::= 'P' | 'U' | 'I'

Prefix    ::= 'm' | 'k' | 'M'

Unit      ::= 'W' | 'V' | 'A'

Additional assertions:

        *The equal sign (`=') will never occur in an other context than within a data eld.

        *There is no whitespace (tabs,blanks) inside a data eld.

        *Either P and U, P and I, or U and I will be given.

 

Output

For each test case, print three lines: a line saying Problem #k" where k is the number of the test case a line giving the solution (voltage, power or current, dependent on what was given), written without a pre x and with two decimal places as shown in the sample output a blank lineACM Contest Problems Archive University of Valladolid (SPAIN）

 

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

该题思路：
   1、从每一行中先找到‘=’字符，
   2、把它之后的数字通过一定办法存储成int型
   3、若后面跟着‘m’‘k’‘M’则相应的对int就行扩大和缩小
   4、根据单位相应的存储到I，P，U中
   5、利用2个参数，求剩下的参数


#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
 
int main ()
{
    int n,j;
    scanf("%dn",&n);
    for (j=1;j<=n;j++)
    {
        char str[100],ch;
        int flag=0,flag1=0,key=0;
        double u=0,i=0,p=0,sum=0;
        double k=0,k1=0.1;
        gets(str);                           //得到字符串
            for (int cou=0;cou<strlen(str);cou++)
            {
                if (str[cou]=='n')
                    break;
                if (str[cou]=='=')       //找到'='
                {
                    flag1=1;
                }
                else if (flag1==1)      //得到数字
                {
                    if (flag==0&&str[cou]>='0'&&str[cou]<='9')
                        k=k*10+str[cou]-'0';
                    else if (str[cou]=='.')
                        flag=1;
                    else if (flag==1&&str[cou]>='0'&&str[cou]<='9')
                    {
                        k=k+(str[cou]-'0')*k1;
                        k1=k1*0.1;
                    }
                    else if (str[cou]=='m')   //单位
                    {
                        k=k/1000;
                    }
                    else if (str[cou]=='M')
                        k = k*1000000;
                    else if (str[cou]=='k')
                        k = k*1000;
                    else if (str[cou]=='A')
                        i=k;
                    else if (str[cou]=='W')
                        p=k;
                    else if (str[cou]=='V')
                        u=k;
                    else
                    {
                        k=0;k1=0.1;flag=0;flag1=0;
                    }
                }
            }
         if (j>1)
                printf("n");
            printf("Problem #%dn",j);   //输出
            //cout<< i << " " << u << " " << p << endl;
            if (u>0&&i>0)
            {
                p = u*i;
                printf("P=%.2lfWn",p);
            }
            else if (u>0&&p>0)
            {
                i = p/u;
                printf("I=%.2lfAn",i);
            }
            else if (p>0&&i>0)
            {
                u = p/i;
                printf("U=%.2lfVn",u);
            }
    }
}