算点程序面向过程与面向对象的物业

3、面向过程的C实现

这是 csdn 算法论坛前版主海星的代码，程序非常简练、精致：

#include

#include

#include

using namespace std;

const double PRECISION = 1E-6;

const int COUNT_OF_NUMBER = 4;

const int NUMBER_TO_BE_CAL = 24;

double number[COUNT_OF_NUMBER];

string expression[COUNT_OF_NUMBER];

bool Search(int n)

{

if (n == 1) {

if ( fabs(number[0] - NUMBER_TO_BE_CAL) PRECISION ) {

cout expression[0] endl;

return true;

} else {

return false;

}

}

for (int i = 0; i n; i++) {

for (int j = i + 1; j n; j++) {

double a, b;

string expa, expb;

a = number[i];

b = number[j];

number[j] = number[n - 1];

expa = expression[i];

expb = expression[j];

expression[j] = expression[n - 1];

expression[i] = \'(\' + expa + \'+\' + expb + \')\';

number[i] = a + b;

if ( Search(n - 1) ) return true;

expression[i] = \'(\' + expa + \'-\' + expb + \')\';

number[i] = a - b;

if ( Search(n - 1) ) return true;

expression[i] = \'(\' + expb + \'-\' + expa + \')\';

number[i] = b - a;

if ( Search(n - 1) ) return true;

expression[i] = \'(\' + expa + \'*\' + expb + \')\';

number[i] = a * b;

if ( Search(n - 1) ) return true;

if (b != 0) {

expression[i] = \'(\' + expa + \'/\' + expb + \')\';

number[i] = a / b;

if ( Search(n - 1) ) return true;

}

将与传统家电厂商合作。 if (a != 0) {

expression[i] = \'(\' + expb + \'/\' + expa + \')\';

number[i] = b / a;

if ( Search(n - 1) ) return true;

}

number[i] = a;

number[j] = b;

expression[i] = expa;

expression[j] = expb;

}

}

return false;

}

void main()

{

for (int i = 0; i COUNT_OF_NUMBER; i++) {

char buffer[20];

int x;

cin x;

number[i] = x;

itoa(x, buffer, 10);

expression[i] = buffer;

}

if ( Search(COUNT_OF_NUMBER) ) {

cout \"Success.\" endl;

} else {

cout \"Fail.\" endl;

}

}

使用任一个 c++ 编译器编译即可。

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