Symbolic Derivation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : Accepted Submission(s) :
Problem Description
Write a program that performs symbolic derivation f锟斤拷(x) = df(x)/dx of a given function f(x). The function f(x) is defined by an expression which may contain the following operations: + (addition), - (subtraction), * (multiplication), / (division), and ln (natural logarithm). Besides, the operands may be the variable x and numerical constants. The expression may contain arbitrarily nested sub-expressions in parentheses ( ). The expression is given in a usual, infix form, such as:
(2*ln(x+1.7)-x*x)/((-7)+3.2*x*x)+(x+3*x)*x
Numerical constants have the form d.d, with an optional sign (+ or -), where the number of digits both in integer and decimal parts are arbitrary. The input expression is guaranteed to be correct (no syntax error can occur).
The output expression should be written in infix form. It should not be optimized for human reading. This means, it can contain redundancies, such as 0*x, 1*x, 0+x, etc. The derivation should be performed using the following rules:
1. The operators * and / are of higher priority than the operators + and -. Parentheses may change the priorities as usually.
2. The operators +, -, *, and / are left-associative, meaning that they group from left to right: a*b*c = (a*b)*c, a/b/c = (a/b)/c, a/b*c = (a/b)*c, etc.
3. The rules for derivation are:
(a + b)' = a' + b'
(a - b)' = a' - b'
(a * b)' = (a' * b + a * b')
(a / b)' = (a' * b - a * b') / b^2 Note: use b^2 and not (b*b) for presentation
ln(a)' = (a')/(a)
x' = 1
const' = 0
4. While producing the symbolic derivation, use parentheses for output strictly as stated in the previous rule. Do not perform presentation optimizations, such as 0*a = 0, 1*a = a, etc.
The input has one f(x) definition per line. The input lines do not have blanks. The output should contain lines with corresponding symbolic derivations f锟斤拷=df/dx, one line for each f. The strings representing f(x) and f 锟斤拷(x) are guaranteed to have no more than 100 characters.
Sample input and output:
(2*ln(x+1.7)-x*x)/((-7)+3.2*x*x)+(x+3*x)*x
Numerical constants have the form d.d, with an optional sign (+ or -), where the number of digits both in integer and decimal parts are arbitrary. The input expression is guaranteed to be correct (no syntax error can occur).
The output expression should be written in infix form. It should not be optimized for human reading. This means, it can contain redundancies, such as 0*x, 1*x, 0+x, etc. The derivation should be performed using the following rules:
1. The operators * and / are of higher priority than the operators + and -. Parentheses may change the priorities as usually.
2. The operators +, -, *, and / are left-associative, meaning that they group from left to right: a*b*c = (a*b)*c, a/b/c = (a/b)/c, a/b*c = (a/b)*c, etc.
3. The rules for derivation are:
(a + b)' = a' + b'
(a - b)' = a' - b'
(a * b)' = (a' * b + a * b')
(a / b)' = (a' * b - a * b') / b^2 Note: use b^2 and not (b*b) for presentation
ln(a)' = (a')/(a)
x' = 1
const' = 0
4. While producing the symbolic derivation, use parentheses for output strictly as stated in the previous rule. Do not perform presentation optimizations, such as 0*a = 0, 1*a = a, etc.
The input has one f(x) definition per line. The input lines do not have blanks. The output should contain lines with corresponding symbolic derivations f锟斤拷=df/dx, one line for each f. The strings representing f(x) and f 锟斤拷(x) are guaranteed to have no more than 100 characters.
Sample input and output:
Input
x*x/x
-45.78*x+x
-2.45*x*x+ln(x-3)
Output
((1*x+x*1)*x-x*x*1)/x^2
(0*x-45.78*1)+1
((0*x-2.45*1)*x-2.45*x*1)+(1-0)/(x-3)
Source
Southeastern Europe 2000
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
using namespace std;
int pri[256];
string str;
string chao(string s)
{
int loc=-1; int lp=0;
char w=' ';
///part one 找优先级最低的运算符
for(int i=s.size()-1;i>=0;i--)
{
switch(s)
{
case')':
lp++; break;
case'(':
lp--; break;
case'+':case'-':
if(i&&((s[i-1]>='0'&&s[i-1]<='9')||s[i-1]=='x'))
{
if(lp==0&&(loc==-1||pri[s]<pri[w]))
{
loc=i;
w=s;
}
}
break;
case'*':case'/':
if(lp==0&&(loc==-1||pri[s]<pri[w]))
{
loc=i;
w=s;
}
break;
case'n':
i--;
if(lp==0&&(loc==-1||pri[s]<pri[w]))
{
loc=i;
w=s;
}
break;
}
}
//cout<<"----> "<<s<<endl;
///patr two 递归输出
string a,b;
switch(w)
{
case'+':case'-':
return chao(s.substr(0,loc))+w+chao(s.substr(loc+1,s.size()-loc-1));
break;
case'*':
a=s.substr(0,loc); b=s.substr(loc+1,s.size()-loc-1);
return "("+chao(a)+"*"+b+"+"+a+"*"+chao(b)+")";
break;
case'/':
a=s.substr(0,loc); b=s.substr(loc+1,s.size()-loc-1);
return "("+chao(a)+"*"+b+"-"+a+"*"+chao(b)+")/"+b+"^2";
break;
case'l':
a=s.substr(3,s.size()-4);
return "("+chao(a)+")/("+a+")";
break;
default:
if(s[0]=='(')
{
return "("+chao(s.substr(1,s.size()-2))+")";
}
else
{
if(s[0]=='x')
return "1";
else
return "0";
}
}
}
int main()
{
pri['+']=pri['-']=1;
pri['*']=pri['/']=2;
pri['l']=3;
string pre;
while(cin>>pre)
{
string str2=chao(pre);
int len=str2.size();
for(int i=0;i<len;i++)
{
if(str2=='+'||str2=='-')
{
if(str2[i+1]=='+'||str2[i+1]=='-')
{
if(str2==str2[i+1])
{
putchar('+');
i++;
}
else
{
putchar('-');
i++;
}
}
else
putchar(str2);
}
else putchar(str2);
}
putchar(10);
}
return 0;
}