检查给定的字符串是否为sum-string

给定字符串数字，请确定它是否为“和字符串”。如果最右边的子字符串可以写为之前两个子字符串的总和，则字符串S称为求和字符串，并且对于它之前的子字符串，递归地适用于此字符串。

例子 ：

“12243660” is a sum string. 
Explanation : 24 + 36 = 60, 12 + 24 = 36

“1111112223” is a sum string. 
Explanation: 111+112 = 223, 1+111 = 112 

“2368” is not a sum string

通常，如果满足以下属性，则将字符串S称为sum-string：

sub-string(i, x) + sub-string(x+1, j) 
 = sub-string(j+1, l)
and 
sub-string(x+1, j)+sub-string(j+1, l) 
 = sub-string(l+1, m) 
and so on till end. 

从示例中，我们可以看到我们的决定取决于前两个选定的数字。
因此，我们为给定的字符串选择所有可能的前两个数字。然后，对于每个选择的两个数字，我们检查它是否为和字符串?因此，方法非常简单。我们使用两个循环，使用两个子串s1和s2生成所有可能的前两个数字。然后检查是否可以使数字s3 =(s1 + s2)。如果我们可以使s3，那么我们递归检查s2 + s3，依此类推。

// C++ program to check if a given string
// is sum-string or not
#include 
using namespace std;
  
// this is function for finding sum of two
// numbers as string
string string_sum(string str1, string str2)
{
    if (str1.size() < str2.size())
       swap(str1, str2);
  
    int m = str1.size();
    int n = str2.size();
    string ans = "";
  
    // sum the str2 with str1
    int carry = 0;
    for (int i = 0; i < n; i++) {
  
        // Sum of current digits
        int ds = ((str1[m - 1 - i] - '0') +
                  (str2[n - 1 - i] - '0') +
                  carry) % 10;
  
        carry = ((str1[m - 1 - i] - '0') +
                 (str2[n - 1 - i] - '0') +
                 carry) / 10;
  
        ans = char(ds + '0') + ans;
    }
  
    for (int i = n; i < m; i++) {
        int ds = (str1[m - 1 - i] - '0' +
                  carry) % 10;
        carry = (str1[m - 1 - i] - '0' +
                 carry) / 10;
        ans = char(ds + '0') + ans;
    }
  
    if (carry)
        ans = char(carry + '0') + ans;
    return ans;
}
  
// Returns true of two substrings of ginven
// lengths of str[beg..] can cause a positive
// result.
bool checkSumStrUtil(string str, int beg,
                     int len1, int len2)
{
  
    // Finding two substrings of given lengths
    // and their sum
    string s1 = str.substr(beg, len1);
    string s2 = str.substr(beg + len1, len2);
    string s3 = string_sum(s1, s2);
  
    int s3_len = s3.size();
  
    // if number of digits s3 is greater then
    // the available string size
    if (s3_len > str.size() - len1 - len2 - beg)
        return false;
  
    // we got s3 as next number in main string
    if (s3 == str.substr(beg + len1 + len2, s3_len)) {
  
        // if we reach at the end of the string
        if (beg + len1 + len2 + s3_len == str.size())
            return true;
  
        // otherwise call recursively for n2, s3
        return checkSumStrUtil(str, beg + len1, len2,
                                              s3_len);
    }
  
    // we do not get s3 in main string
    return false;
}
  
// Returns true if str is sum string, else false.
bool isSumStr(string str)
{
    int n = str.size();
  
    // choosing first two numbers and checking
    // whether it is sum-string or not.
    for (int i = 1; i < n; i++)
        for (int j = 1; i + j < n; j++)
            if (checkSumStrUtil(str, 0, i, j))
                return true;
  
    return false;
}
  
// Driver code
int main()
{
    cout << isSumStr("1212243660") << endl;
    cout << isSumStr("123456787");
    return 0;
}

输出：

1
0