Question

C语言:1到13中任意抽取4个数字(可重复),通过加减乘除计算得到24,要求输出所有算式:无法计算出24则输出无

C语言:1到13中任意抽取4个数字(可重复),通过加减乘除计算得到24,要求输出所有算式:无法计算出24，则输出无解

Answer 1

#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 false;
} 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;
}

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;
}
}

追问

这个完全没用啊

Answer 2

http://zhidao.baidu.com/question/21809714.html?si=9