1 枚舉，沒什么難度，這個可以做為筆試題來出了。。

do

{

// +-*/   4*4*4

}while(next_permutation())

復雜度是4*4*4*4!

2 遞歸求解  網上的一個解法，找不到出處了。。。

#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)
{
    /*n==1表示一次計算結束，number[0]中即為計算的結果*/
    if (n == 1)
    {
        if ( fabs(number[0] - NUMBER_TO_BE_CAL) < PRECISION )
        {
            /*expression[0]中保存了求解過程*/
            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];
            /*將剩下的有效數字往前挪，
             *由于兩數計算結果保存在number[i]中，
             *所以將數組末元素覆蓋number[j]即
            */
            number[j] = number[n - 1];
            expa = expression[i];
            expb = expression[j];
            expression[j] = expression[n - 1];
            /*計算a+b*/
            expression[i] = '(' + expa + '+' + expb + ')';
            number[i] = a + b;
            if ( Search(n - 1) ) return true;
            /*計算a-b*/
            expression[i] = '(' + expa + '-' + expb + ')';
            number[i] = a - b;
            if ( Search(n - 1) ) return true;
            /*計算b-a*/
            expression[i] = '(' + expb + '-' + expa + ')';
            number[i] = b - a;
            if ( Search(n - 1) ) return true;
            /*計算(a*b)*/
            expression[i] = '(' + expa + '*' + expb + ')';
            number[i] = a * b;
            if ( Search(n - 1) ) return true;
            /*計算(a/b)*/
            if (b != 0)
            {
                expression[i] = '(' + expa + '/' + expb + ')';
                number[i] = a / b;
                if ( Search(n - 1) ) return true;
            }
            /*計算(b/a)*/
            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;
}

int 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;
    return 0;
}