广州市网站建设品牌,聊城做网站费用价格,郑州app开发定制多少钱,北京网站优化招聘sql 算出下级销售总和Description: 描述#xff1a; 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#xff0c;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…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 T3
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(j1,k) str.substring(k1,l)
str.substring(j1,k) str.substring(k1,l) str.substring(l1,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 posij 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
翻译自: https://www.includehelp.com/icp/find-out-the-sum-string.aspxsql 算出下级销售总和
