c语言编程求出加减乘除运算符。具体问题看问题补充。谢谢。

已知一个数，如10，并且还已知用3加减乘除五次等于这个10，其关系表达式为：3*3-3+3+3/3=10，其中加减乘除运算符为* - + + /
另换一个数，如43，它的运算数为7，并且加减乘除一共六次。问：如何用c语言编程求出这些运算符？
注解：那些运算符我都知道（不止一组可能），我主要是要的程序源代码。跪求了！！
运算符中间不可以用括号

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推荐答案 2013-02-23

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// Ex6_10Extended.cpp
// A program to implement a calculator accepting parentheses
#include <iostream> // For stream input/output
#include <cstdlib> // For the exit() function
#include <cctype> // For the isdigit() function
#include <cstring> // For the strcpy() function
using std::cin;
using std::cout;
using std::endl;
void eatspaces(char* str); // Function to eliminate blanks
double expr(char* str); // Function evaluating an expression
double term(char* str, int& index); // Function analyzing a term
double number(char* str, int& index); // Function to recognize a number
char* extract(char* str, int& index); // Function to extract a substring
const int MAX(80); // Maximum expression length,
// including '\0'
int main()
{
char buffer[MAX] = {0}; // Input area for expression to be evaluated
cout << endl
<< "Welcome to your friendly calculator."
<< endl
<< "Enter an expression, or an empty line to quit."
<< endl;
for(;;)
{
cin.getline(buffer, sizeof buffer); // Read an input line
eatspaces(buffer); // Remove blanks from input
if(!buffer[0]) // Empty line ends calculator
return 0;
cout << "\t= " << expr(buffer) // Output value of expression
<< endl << endl;
}
}
// Function to eliminate spaces from a string
void eatspaces(char* str)
{
int i(0); // 'Copy to' index to string
int j(0); // 'Copy from' index to string
while((*(str + i) = *(str + j++)) != '\0') // Loop while character
// copied is not \0
if(*(str + i) != ' ') // Increment i as long as
i++; // character is not a space
return;
}
// Function to evaluate an arithmetic expression
double expr(char* str)
{
double value(0.0); // Store result here
int index(0); // Keeps track of current character position
value = term(str, index); // Get first term
for(;;) // Indefinite loop, all exits inside
{
switch(*(str + index++)) // Choose action based on current character
{
case '\0': // We're at the end of the string
return value; // so return what we have got
case '+': // + found so add in the
value += term(str, index); // next term
break;
case '-': // - found so subtract
value -= term(str, index); // the next term
break;
default: // If we reach here the string
cout << endl // is junk
<< "Arrrgh!*#!! There's an error"
<< endl;
exit(1);
}
}
}
// Function to get the value of a term
double term(char* str, int& index)
{
double value(0.0); // Somewhere to accumulate
// the result
value = number(str, index); // Get the first number in the term
// Loop as long as we have a good operator
while(true)
{
if(*(str + index) == '*') // If it's multiply,
value *= number(str, ++index); // multiply by next number
else if(*(str + index) == '/') // If it's divide,
value /= number(str, ++index); // divide by next number
else
break;
}
return value; // We've finished, so return what
// we've got
}
// Function to recognize a number in a string
double number(char* str, int& index)
{
double value(0.0); // Store the resulting value
if(*(str + index) == '(') // Start of parentheses
{
char* psubstr(nullptr); // Pointer for substring
psubstr = extract(str, ++index); // Extract substring in brackets
value = expr(psubstr); // Get the value of the substring
delete[]psubstr; // Clean up the free store
return value; // Return substring value
}
// There must be at least one digit...
if(!isdigit(*(str + index)))
{ // There's no digits so input is junk...
cout << endl
<< "Arrrgh!*#!! There's an error"
<< endl;
exit(1);
}
while(isdigit(*(str + index))) // Loop accumulating leading digits
value = 10*value + (*(str + index++) - '0');
// Not a digit when we get to here
if(*(str + index) != '.') // so check for decimal point
return value; // and if not, return value
double factor(1.0); // Factor for decimal places
while(isdigit(*(str + (++index)))) // Loop as long as we have digits
{
factor *= 0.1; // Decrease factor by factor of 10
value = value + (*(str + index) - '0')*factor; // Add decimal place
}
return value; // On loop exit we are done
}
// Function to extract a substring between parentheses
// (requires cstring)
char* extract(char* str, int& index)
{
char buffer[MAX]; // Temporary space for substring
char* pstr(nullptr); // Pointer to new string for return
int numL(0); // Count of left parentheses found
int bufindex(index); // Save starting value for index
do
{
buffer[index - bufindex] = *(str + index);
switch(buffer[index - bufindex])
{
case ')':
if(0 == numL)
{
size_t size = index - bufindex;
buffer[index - bufindex] = '\0'; // Replace ')' with '\0'
++index;
pstr = new char[index - bufindex];
if(!pstr)
{
cout << "Memory allocation failed,"
<< " program terminated.";
exit(1);
}
strcpy_s(pstr, index-bufindex, buffer); // Copy substring to new memory
return pstr; // Return substring in new memory
}
else
numL--; // Reduce count of '(' to be matched
break;
case '(':
numL++; // Increase count of '(' to be
// matched
break;
}
} while(*(str + index++) != '\0'); // Loop - don't overrun end of string
cout << "Ran off the end of the expression, must be bad input."
<< endl;
exit(1);
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