Artificial Intelligence?(字符串处理)

Artificial Intelligence?

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

  Artificial Intelligence? 

Physics teachers in high school often think that problems given as text are more demanding than pure computations. After all, the pupils have to read and understand the problem first!

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 first. 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 first line of the input file will contain the number of test cases.

Each test case will consist of one line containing exactly two data fields and some additional arbitrary words. A data field 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 prefixes m (milli), k (kilo) and M (Mega) may also occur. To summarize it: Data fields adhere to the following grammar:

DataField ::= Concept '=' RealNumber [Prefix] UnitConcept   ::= '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 field.
• There is no whitespace (tabs,blanks) inside a data field.
• 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 prefix and with two decimal places as shown in the sample output
• a blank line

Sample Input 

3If 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 #1P=900.00WProblem #2I=0.45AProblem #3U=1250000.00V

这道题是给你一句话，其中会包含P,U,I中的两个，让你根据公式求出P=U*I 求出第三个的值并输出。。。

在处理的时候要注意它们的单位，MW（1000000W），kV（1000V），mA（0.001A），然后还要注意这三个的值可能为实数。。

本来我是直接输入每一句话，找到=后判断前后的字符，发现这样不好表示，尤其是在=后面为实数的时候，其实输入的数据也就=前后有用，所以我直接用double保存=后的实数，这样可以直接得到两个实数

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int N = 1500;int main(){    int T;    cin>>T;    for(int r=1; r<=T; r++)    {        char a[N];        int sa=0;        while(cin>>a[sa])        {            if(a[sa]=='=')                break;            sa++;        }        double x;        char tx;        cin>>x;        cin>>tx;        int fp=1,fu=1,fi=1;        if(tx=='M')            x*=1000000;        if(tx=='m')            x*=0.001;        if(tx=='k')            x*=1000;        if(a[sa-1]=='P')            fp=0;        else if(a[sa-1]=='U')            fu=0;        else if(a[sa-1]=='I')            fi=0;        char b[N];        int sb=0;        while(cin>>b[sb])        {            if(b[sb]=='=')                break;            sb++;        }        double y;        char ty;        cin>>y;        cin>>ty;        int rp=1,ru=1,ri=1;        if(ty=='M')            y*=1000000;        else if(ty=='m')            y*=0.001;        else if(ty=='k')            y*=1000;        if(b[sb-1]=='P')            rp=0;        else if(b[sb-1]=='U')            ru=0;        else if(b[sb-1]=='I')            ri=0;        printf("Problem #%d\n",r);        if(!fp && !ri)            printf("U=%.2lfV\n",x/y);        if(!fp && !ru)            printf("I=%.2lfA\n",x/y);        if(!fu && !rp)            printf("I=%.2lfA\n",y/x);        if(!fu && !ri)            printf("P=%.2lfW\n",x*y);        if(!fi && !rp)            printf("U=%.2lfV\n",y/x);        if(!fi && !ru)            printf("P=%.2lfW\n",x*y);        printf("\n");    }    return 0;}