UVA 537 (13.08.01)

 Artificial Intelligence? 

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 first!

So they don't state a problem like ``U=10V, I=5A, P=?" but rather like ``You have an electricalcircuit that contains a battery with a voltage of U=10V and a light-bulb. There's an electricalcurrent 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 thetext 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 atleast 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 theP-U-I type problems first. That means, problems in which two of power, voltage and current aregiven and the third is wanted.

Your job is to write a program that reads such a text problem and solves it according to the simplealgorithm 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 additionalarbitrary words. A data field will be of the formI=xA,U=xV or P=xW, wherex is a real number.

Directly before the unit (A, V or W) one of the prefixesm (milli),k (kilo) and M (Mega) may alsooccur. 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 andI 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), writtenwithout 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

题意: 给出一段话, 包含了电流I, 电压U, 或者电功率P的数值, 要我们判断出到底是给出了哪两个数值, 并求另一个数值

做法: 可能一开始会没有头绪, 但是仔细一想, 切入点应该是 = 号, 遇到等号后面会跟着一个数值, 那么我们就可以在遇到 = 号的时候, scanf数值~

然后, 重点又来了, 数值取完后, 后面可能是m, k, 或 M等, 而不是单位

我的做法是, 先getchar(), 如果是m, k, 或M的其中一个, 那么对数值进行处理一下, 继续getchar(), 不然的话, 再判断是否是A, V, W等单位

那最后一步就更简单了,根据单位是什么就可以判定是把数值num给I还是U还是P~

AC代码:

#include<stdio.h>int main() {int N, cas = 0;int mark;double I, P, U;double num;char ch;scanf("%d%c", &N, &ch);while(N--) {mark = num = I = P = U = 0;while(ch = getchar()) {if(ch == '\n')break;if(ch == '=' && mark < 2) {scanf("%lf", &num);mark++;ch = getchar();if(ch == 'm') {num = num / 1000;ch = getchar();}else if(ch == 'k') {num = num * 1000;ch = getchar();}else if(ch == 'M') {num = num * 1000000;ch = getchar();}if(ch == 'A')I = num;else if(ch == 'V')U = num;else if(ch == 'W')P = num;}}printf("Problem #%d\n", ++cas);if(P && I)printf("U=%.2lfV\n", P/I);else if(I && U)printf("P=%.2lfW\n", I*U);else if(P && U)printf("I=%.2lfA\n", P/U);printf("\n");}return 0;}