趣味算法-巧填运算符

给定一个整数数组，和一个单独的数字，在数组的每一个元素中间填加 "+"或"-" 使其运算结果等于单独的数字例如给定的数组为{7 2 4} ，数字为 9。运算结果为7-2+4=9

规则1：数组中元素的位置不能变化。

规则2：如果无法实现则输出 Invalid

举例：

Input:

1 2 3 4 10

1 2 3 4 5

Output:

1+2+3+4=10

Invalid

想法：使用穷举法，使用递归在每个位置尝试每个运算符，如果不成立，则需要返回尝试前的状态。

程序示例：（给的测试数字中，最后一个为单独的结果）

#include <iostream>

using namespace std;

void calc(int arr[], int length, int final)
{
static int index = 0;
static int sum = arr[0];
static bool isfind = false, iscomplete = false;
static char* operators = NULL;

if (index == 0)
{
operators = new char[length];
memset(operators, '+', length);
}

if (isfind)
return;

if (index == length-1)
{
if (sum == final)
{
iscomplete = true;
isfind = true;

operators[length-1] = '=';
for (int i = 0; i < length; i++)
cout << arr[i] << operators[i]; //ok
cout << final << endl;
delete[] operators;
operators = NULL;
}
else
{
if (index == 0)
{
delete[] operators;
operators = NULL;
cout<<"Invalid" << endl;
}
}
return;
}

if (!isfind)
{
index++;
sum += arr[index];
operators[index] = '+';

calc(arr, length, final);

if ((!isfind) && (!iscomplete))
{
sum -= arr[index];
operators[index] = '+';
index--;
}
}

if (!isfind)
{
index++;
sum -= arr[index];
operators[index] = '-';
calc(arr, length, final);

if ((!isfind) && (!iscomplete))
{
sum += arr[index];
operators[index] = '+';
index--;
}
}

if ((index == 0) && (!isfind))
{
delete[] operators;
operators = NULL;
cout<<"Invalid" << endl;
}
}

int main()
{
//int arr[2] = {3, 3};
//int arr[2] = {3, 4};
//int arr[3] = {2, 3, 4};
int arr[4] = {7, 2, 4, 9};
//int arr[5] = {1, 2, 3, 4, 10};
int length = sizeof(arr)/sizeof(arr[0]);

calc(arr, length-1, arr[length-1]);

cout<<endl;

cin >> length;
return 0;
}