用"-“符号计算数学表达式失败

Question

我正在创建一个评估数学表达式的程序。它需要两个输入:一个整数和一个空间分隔的数学表达式的字符串。它应该返回与整数输入匹配的表达式的索引。

样本输入：

20
(2+2) (10+10) (3*5*6)

样本输出：

index 1

似乎我的逻辑在使用减法运算符时失败了，所以我集中在代码的“减法评估”部分。所有其他运营商似乎都成功了。我想弄清楚我错过了什么。任何帮助都是非常感谢的-谢谢。

失败的一个示例输入是：

-8
(2-2-2-2-2-2)

这是我的代码：

#include <iostream>
#include <algorithm>
#include <regex>
using namespace std;

int main() {
    
 int input;
 string w;
    
 cin >> input;
    
 // ws ignore spaces or "whitespace"
 getline(cin >> ws, w);
    
 int w_len = w.length();
    
 // count total expressions in string
 int exp = 1;
 for (int i = 0; i <= w_len; i++) {
     if (isspace(w[i])) {
         exp++;
     }
 }
    
 // create array to hold expressions
 string w_arr[exp];
    
 // populate s_arr elements with the expressions
 int wcount = 0;
 for (int i = 0; i <= w_len; i++) {
     if (w[i] == ' ') {
         wcount++;
         i++; 
     }
     w_arr[wcount] += w[i];
 }
    
 int w_size = sizeof(w_arr) / sizeof(w_arr[1]);
    
 string s;
 int index = -1;
 
 // start of the evaluation loop
 for (int k = 0; k < w_size; k++) {
 index++;
 s = w_arr[k];
    
    
 s.erase(remove(s.begin(),s.end(),'('), s.end());
 s.erase(remove(s.begin(),s.end(),')'), s.end());
    
    // count total punctuation
    int count = 0;
    for (size_t i = 0; i < s.length(); i++) {
        if (ispunct(s[i])) {
            count++;
        }
    }
    
    // add spaces around operators
    s = regex_replace(s, regex("\\+"), " + ");
    s = regex_replace(s, regex("\\*"), " * ");
    s = regex_replace(s, regex("\\/"), " / ");
    s = regex_replace(s, regex("\\-"), " - ");
    
    int count2 = 0;
    int size = count + count + 1;
    string z[size];
    
    // create array z[] from input string
    for (size_t i = 0; i < s.length(); i++) {
        if (s[i] == ' ') {
            count2++;
            i++;
        }
        z[count2] += s[i];
    }
    
    // multiplication evaluation
    string res;
    mulStart:
    for (int j = 0; j <= size; j++) {
        if (z[j] == "*") {
            int mul1 = stoi(z[j-1]);
            int mul2 = stoi(z[j+1]);
            int mul = mul1 * mul2;
            res = to_string(mul);
            
            // overwrite expression with result
            z[j-1] = res;
            for (int i = j; i<=size-3; i++) {
                z[i] = z[i+2];
            }
            
            // replace end of array with blanks
            for (int i = size-2; i<size; i++) {
                z[i] = " ";
            }
        }
    }
    
    // check for other mul operators
    size = sizeof(z) / sizeof(z[0]);
    int mulCount = 0;
    for (int i = 0; i <= size; i++) {
        if (z[i] == "*") {
            mulCount++;
        }
    }
    if (mulCount > 0) {
        goto mulStart;
    }
    
    // division evaluation
    divStart:
    for (int j = 0; j <= size; j++) {
        if (z[j] == "/") {
            int div1 = stoi(z[j-1]);
            int div2 = stoi(z[j+1]);
            int div = div1 / div2;
            res = to_string(div);
            
            // overwrite expression with result
            z[j-1] = res;
            for (int i = j; i<=size-3; i++) {
                z[i] = z[i+2];
            }
            
            // replace end of array with blanks
            for (int i = size-2; i<size; i++) {
                z[i] = " ";
            }
        }
    }
    
    // check for other div operators
    size = sizeof(z) / sizeof(z[0]);
    int divCount = 0;
    for (int i = 0; i <= size; i++) {
        if (z[i] == "/") {
            divCount++;
        }
    }
    if (divCount > 0) {
        goto divStart;
    }
    
    // addition evaluation
    addStart:
    for (int j = 0; j <= size; j++) {
        if (z[j] == "+") {
            int add1 = stoi(z[j-1]);
            int add2 = stoi(z[j+1]);
            int add = add1 + add2;
            res = to_string(add);
            
            // overwrite expression with result
            z[j-1] = res;
            for (int i = j; i<=size-3; i++) {
                z[i] = z[i+2];
            }
            
            
            // replace end of array with blanks
            for (int i = size-2; i<size; i++) {
                z[i] = " ";
            }
        }
    }
    
    // check for other add operators
    size = sizeof(z) / sizeof(z[0]);
    int addCount = 0;
    for (int i = 0; i <= size; i++) {
        if (z[i] == "+") {
            addCount++;
        }
    }
    if (addCount > 0) {
        goto addStart;
    }
    
    // subtraction evaluation
    subStart:
    for (int j = 0; j <= size; j++) {
        if (z[j] == "-") {
            int sub1 = stoi(z[j-1]);
            int sub2 = stoi(z[j+1]);
            int sub = sub1 - sub2;
            res = to_string(sub);
            
            // overwrite expression with result
            z[j-1] = res;
            for (int i = j; i<=size-3; i++) {
                z[i] = z[i+2];
            }
            
            // replace end of array with blanks
            for (int i = size-2; i<size; i++) {
                z[i] = " ";
            }
        }
    }
    
    // check for other sub operators
    size = sizeof(z) / sizeof(z[0]);
    int subCount = 0;
    for (int i = 0; i <= size; i++) {
        if (z[i] == "-") {
            subCount++;
        }
    }
    if (subCount > 0) {
        goto subStart;
    }
     
    
    // print out index result
    if (stoi(z[0]) == input) {
        cout << "index " << index << endl;
        break;
    }
    
    // print "none" otherwise
    if (k == w_size - 1) {
        cout << "none" << endl;
    }
    
 }
 
    return 0;
}

Answer 1

"//检查其他子操作符“，这应该给您一个提示。您已经迭代了所有的输入，但似乎有"-“之后，您需要再次这样做。但你可以计算出"-“无序。

问题是，即使在找到一个j时，也可以提前索引-。通过消除一个-，下一个-取代您删除的那个。然后你就跳过了。

让我们以您为例，我对-进行了编号，以说明它们在循环时是如何移动的：

"2-2-2-2-2-2" becomes:

"2" "-"1 "2" "-"2 "2" "-"3 "2" "-"4 "2" "-"5 "2"
 ^

"2" "-"1 "2" "-"2 "2" "-"3 "2" "-"4 "2" "-"5 "2"
     ^

"0" "-"2 "2" "-"3 "2" "-"4 "2" "-"5 "2"
          ^

"0" "-"2 "2" "-"3 "2" "-"4 "2" "-"5 "2"
              ^ 

"0" "-"2 "0" "-"4 "2" "-"5 "2"
                   ^ 

"0" "-"2 "0" "-"4 "2" "-"5 "2"
                       ^ 

"0" "-"2 "0" "-"4 "0"

然后你去找subStart。

同样的问题应该出现在/上。

越来越糟的是，在*之前，您正在做/。那么4 * 9 / 2 * 3会发生什么呢？+和-：1 + 2 - 3 + 4也是如此。