题目链接:http://poj.org/problem?id=2256
Artificial Intelligence?
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1323 | Accepted: 643 |
Description
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 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] 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 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
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
Source
题意:给定一个文本，提取其中的已知条件，根据公式计算结果。
题解:字符串模拟题。题目描述中已经说明等于号不会出现在除"已知条件"外其他的地方，所以可以根据这个来把两个已知条件抠出来。提取出来后，把里面的字符形式的数字转换成double型的数字。已知两个条件计算第三个条件便是。这种题目没有什么特定解法，主要是要有熟悉字符串并且考虑全面。具体细节实现看代码。
AC代码:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
using namespace std;
string text, Map = "PUI", pMap = "mkM",out="WVA";
int t; double res[3];
int Hash(char c,string x){
for (int i = 0; i<3; i++)
if (x[i] == c)return i; return -1;
}
double pHash(char c){
switch (c){
case 'm':return 0.001;
case 'k':return 1000;
case 'M':return 1000000;
}
}
string deal(string s, int flag){
int len = s.size(), b, e;
string tmpstr;
if (flag){//第二个等号
for (int i = len - 1; i > 0; i--)
if (s[i] == '='){
b=i-1;
for(int j=i+1;j<len;j++)
if(Hash(s[j],out)!=-1||j==len-1){
e=j;break;
}
tmpstr = s.substr(b, e - b + 1);
break;
}
}
else{ //第一个等号
for (int i = 0; i<len; i++)
if (s[i] == '='){
b=i-1;
for(int j=i+1;j<len;j++)
if(Hash(s[j],out)!=-1||j==len-1){
e=j;break;
}
tmpstr = s.substr(b, e - b + 1); break;
}
}
return tmpstr;
}
double ToNum(string s){ //把数字抠出来
double sum = 0;
int len = s.size(), flag = 0;
for (int i = 2; i<len - 1; i++){
if (s[i] == '.'){
flag = -1; continue;
}
if (s[i] >= 'A'&&s[i] <= 'Z'||s[i]>='a'&&s[i]<='z'){
sum *= pHash(s[i]); break;
}
if (flag)sum += (s[i] - '0')*pow(10.0, flag), flag--;
else sum = (s[i] - '0')+sum*10;
}
return sum;
}
double calc(int k){
switch (k){
case 0:return res[1] * res[2];
case 1:return res[0] / res[2];
case 2:return res[0] / res[1];
}
}
int main()
{
cin.sync_with_stdio(false);
cin >> t; cin.ignore(); //吃掉那个回车，这个不能少
for (int k = 1; k <= t; k++){
cout << "Problem #" << k << endl;
int flag[3] = { 0 };
getline(cin, text);
string s1 = deal(text, 0),s2 = deal(text, 1);
int pos1 = Hash(s1[0],Map), pos2 = Hash(s2[0],Map);
flag[pos1] = 1; flag[pos2] = 1;
res[pos1] = ToNum(s1); res[pos2] = ToNum(s2);
for (int i = 0; i<3; i++)
if (!flag[i]){
printf("%c=%.2lf%c\n\n", Map[i], calc(i),out[i]); break;
}
}
return 0;
}
【转载请注明出处】
作者:MummyDing
出处:http://blog.csdn.net/mummyding/article/details/43818357