24点算法，送分哦？

[forrest_wang]

我现在在做一个24点的程序，但就是算法不好，太复杂了，谁有好的算法啊，提供来参考一下，有分哦！

[xuying]

给你的源程序看看，用递归。
#include <iostream>
#include <string>
#include <cmath>

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;
}
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;
}
}

[forrest_wang]

楼上的兄弟，谢了，算法不错，和我想的很象。大家都来啊，还有更好的都拿出来啊！！

[bigcgq]

各位有没有看过95年的软件报合订本，那里头有的。嘿嘿........

[forrest_wang]

我没有这本书啊，传上来看看啊

[forrest_wang]

怎么就没有了！！！

[OpenVMS]

5年前做过个,递归分解求表达式