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(wǎng)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(shù)字往前挪，
*由于兩數(shù)計算結果保存在number[i]中，
*所以將數(shù)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;
}