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;}