sql 算出下级销售总和_找出总和字符串

sql 算出下级销售总和

Description:

描述：

This is a standard interview problem to check that the given string is a sum string or not using backtracking.

这是一个标准的面试问题，用于检查给定的字符串是否为总和字符串或不使用回溯。

Problem statement:

问题陈述：

There is a string given to you. You have to find out whether the given string is a sum string or not. For a string "12345168213" if it goes like this, "123" + "45" = "168" "45" + "168" = "213" then it is a sum string otherwise not.

有一个字符串给你。您必须找出给定的字符串是否为求和字符串。对于字符串“ 12345168213”，如果这样， “ 123” + “ 45” = “ 168” “ 45” + “ 168” = “ 213” ，则为求和字符串，否则为非。

    Input:
Test case T
T no. of strings we have to check. 
E.g.
3
"12345168213"
"123581321"
"15643"
Output:
If the string is a sum-string then you have to print 
"It is a sum string"
and, if it is not then you have to print 
"It is not a sum string".

Example

例

    T=3
Str= "12345168213"
"123" + "45" = "168"
"45" + "168" = "213"
It is a sum string
Str= "123581321" 
"1" + "2" = "3"
"2" + "3" = "5"
"3" + "5" = "8"
"5" + "8" = "13"
"8" + "13" = "21"
It is a sum string
Str= "15643"
"1" + "5" = "6"
"5" + "6" = "11"
It is not a sum string

Explanation with example

举例说明

To find out the actual length of the first two substrings is a problem of combination. We will solve the problem using the backtracking process. If we break the problem into sub-problems like,

找出前两个子串的实际长度是组合的问题。我们将使用回溯过程解决问题。如果我们将问题分解为子问题，

Find out the length of the first sub-string.
找出第一个子字符串的长度。
Find out the length of the second sub-string.
找出第二个子字符串的长度。
Add the two strings.
添加两个字符串。
Check if the summation is the same as the next substring.
检查求和是否与下一个子字符串相同。
If yes, we will continue the process otherwise the string is not a sum string.
如果是，我们将继续执行该过程，否则该字符串不是求和字符串。

    str.substring(i,j) + str.substring(j+1,k) = str.substring(k+1,l)
str.substring(j+1,k) + str.substring(k+1,l) = str.substring(l+1,m)

Where, i, j, k, l, m is the position index on the substring of the string.

其中， i ， j ， k ， l ， m是字符串的子字符串的位置索引。

For the string "12345168213"

对于字符串“ 12345168213”

To find out the length of the first substring we will look for the all possible combination of the sub-string of the string from the starting index.

为了找出第一个子字符串的长度，我们将从起始索引中查找该字符串的子字符串的所有可能组合。

"1" , "12" , "123" , "1234" , "12345" , "123451" etc.

“ 1” ， “ 12” ， “ 123” ， “ 1234” ， “ 12345” ， “ 123451”等

To find out the length of the second substring we will look for all possible combinations of the sub-string of the string from the last index of the first substring.

为了找出第二个子字符串的长度，我们将从第一个子字符串的最后一个索引中查找该字符串的子字符串的所有可能组合。

Let the first index= "123" then the all possible combinations are "4", "45", "451", "4516" etc.

假设第一个索引= “ 123”，那么所有可能的组合都是“ 4” ， “ 45” ， “ 451” ， “ 4516”等。

After calculating the summation of the two sub-strings will also find out the length of the result and take the next sub-string who has a length equal to that length.

在计算完两个子字符串的总和后，还将找出结果的长度，并获取下一个长度等于该长度的子字符串。

If the two substrings are "123" and "45" after calculating the summation the result "168" we will calculate its length which is equal to 3 and take the next sub-string e.g. "168"

如果两个子字符串分别是“ 123”和“ 45” ，计算出结果“ 168”后，我们将计算其长度等于3，并取下一个子字符串，例如“ 168”

Both the resultant and the sub-string are the same therefore we will add up "45" and "168" and check with "213".

结果和子字符串都相同，因此我们将“ 45”和“ 168”加起来并用“ 213”进行校验。

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
//adding the numbers
string add(string str1, string str2)
{
int len1 = str1.length();
int len2 = str2.length();
int i = 1;
int carry = 0;
string str = "";
while ((len1 - i) >= 0 && (len2 - i) >= 0) {
int result = (str1[len1 - i] - '0') + (str2[len2 - i] - '0') + carry;
char ch = (result % 10) + '0';
str = ch + str;
carry = result / 10;
i++;
}
while ((len1 - i) >= 0) {
int result = (str1[len1 - i] - '0') + carry;
char ch = (result % 10) + '0';
str = ch + str;
carry = result / 10;
i++;
}
while ((len2 - i) >= 0) {
int result = (str2[len2 - i] - '0') + carry;
char ch = (result % 10) + '0';
str = ch + str;
carry = result / 10;
i++;
}
if (carry > 0) {
char ch = carry + '0';
str = ch + str;
}
return str;
}
bool checksumUtill(string str, int pos, int str1_len, int str2_len)
{
int n = str.length();
int i = str1_len, j = str2_len;
//if the total sum of the current position and
//both the strings is greater than total length
if (pos + str1_len + str2_len >= n)
return false;
//calculate the sum
string s = add(str.substr(pos, i), str.substr(pos + i, j));
//if the next substring of pos+i+j is equals to the sum
if (s == str.substr(pos + i + j, s.length())) {
if (pos + i + j + s.length() == n)
return true;
return checksumUtill(str, pos + i, j, s.length());
}
return false;
}
bool check_sum_string(string str)
{
if (str.length() == 0)
return false;
int str1_len = 1;
int str2_len = 1;
int pos = 0;
//go for all the combinations
for (int i = 1; i < str.length(); i++) {
for (int j = 1; j < str.length(); j++) {
if (checksumUtill(str, pos, i, j)) {
return true;
}
}
}
return false;
}
int main()
{
int t;
cout << "Test Case : ";
cin >> t;
while (t--) {
string str;
cout << "Enter the String : ";
cin >> str;
if (check_sum_string(str)) {
cout << "It is a sum string\n";
}
else {
cout << "It is not a sum string\n";
}
}
return 0;
}

Output

输出量

Test Case : 3
Enter the String : 12345168213
It is a sum string
Enter the String : 123581321
It is a sum string
Enter the String : 15648
It is not a sum string