所有子序列的逆序对总和
Problem statement:
问题陈述：
Given an integer, S represented as a string, get the sum of all possible substrings of this string.
给定一个以字符串形式表示的整数S ，得到该字符串所有可能的子字符串的和 。
Input:
输入：
A string S that representing the number.
代表数字的字符串S。
Output:
输出：
Print sum of all possible substrings as required result.
根据要求的结果打印所有可能的子字符串的总和。
Constraints:
限制条件：
1 <= T <= 100
1 <= S <= 1012
Example:
例：
Input:
1234
326
Output:
1670
395
Explanation:
说明：
For the first input 1234,
All possible substrings are
1, 2, 3, 4, 12, 13, 23, 34, 123, 234, 1234
Total sum = 1 + 2 + 3 + 4 + 12 + 23 + 34 + 123 + 234 + 1234 = 1670
For the second input 326
All possible substrings are
3, 2, 6
32, 26
326
Total sum=3+2+6+32+26+326= 395
Solution Approach:
解决方法：
The solution approach is by storing the substring sums to compute the exact next substring sum
解决方法是通过存储子字符串和以计算确切的下一个子字符串和
Create dp[n][n] to store substring sums;
创建dp [n] [n]来存储子字符串和；
Initialize sum=0 which will be our final result;
初始化sum = 0，这将是我们的最终结果；
Base case computation (single length substrings),
基本案例计算(单长度子字符串)，
for i=0 to n-1,n= string length dp[i][i]=s[i] -'0'; //s[i]-'0' gives the digit actually sum+=dp[i][i]; end forTill now we have computed all single digit substrings,
到现在为止，我们已经计算了所有个位数的子字符串，
for substring length,len=2 to n for start=0 to n-len //so basically it's the substring s[start,end] int end=start+len-1; dp[start][end]=dp[start][end-1]*10+s[end]-'0'; sum+=dp[start][end]; end for end forSum is the final result.
总和是最终结果。
All the statements are self-explanatory except the one which is the fundamental idea of the entire storing process. That is the below one,
所有陈述都是不言自明的，只是整个存储过程的基本思想。 那是下一个，
dp[start][end]=dp[start][end-1]*10+s[end]-'0';
Let's check this with an example,
我们来看一个例子，
Say we are computing for string s="1234"
At some stage of computing,
Start=1, end= 3
So
Dp[start][end]=dp[start][end-1]*10+s[end]-'0'
So basically we are computing value of substring s[start..end]
with help of already computed s[start,end-1]
For this particular example
s[start..end] ="234"
s[start..end-1] ="23"
Now, dp[1][3]=dp[1][2]*10+'4'-'0'
So, assuming the fact that our algo is correct and thus dp[start][end-1]
has the correct value, dp[]1[2] would be 23 then
So,
dp[1][3]=23*10+'4'-'0=234
and that's true
So, here's the main logic
Now how dp[1][2] is guaranteed to be correct can be
explored if we start filling the Dp table from the base conditions?
Let's start for the same example
让我们开始同样的例子
N=4 here
N = 4这里
So, we need to fill up a 4X4 DP table,
因此，我们需要填写4X4 DP表，
After filling the base case,
装完基本外壳后，
Now, I am computing for len=2
现在，我正在计算len = 2
Start=0, end=1
开始= 0，结束= 1
Start=1, end=2
开始= 1，结束= 2
Start=2, end=3
开始= 2，结束= 3
For len =3
对于len = 3
Start=0, end=2
开始= 0，结束= 2
Start=1, end=3
开始= 1，结束= 3
Len=4
Len = 4
Start=0, end=3
开始= 0，结束= 3
At each step we have summed up, so result is stored at sum.
在每一步我们都进行了总结，因此结果被存储在总和中。
C++ Implementation:
C ++实现：
#include <bits/stdc++.h>
using namespace std;
void print(vector<int> a, int n)
{
for (int i = 0; i < n; i++)
cout << a[i] << " ";
cout << endl;
}
long long int my(string s, int n)
{
long long int dp[n][n];
long long int sum = 0;
for (int i = 0; i < n; i++) {
dp[i][i] = s[i] - '0';
sum += dp[i][i];
}
for (int len = 2; len <= n; len++) {
for (int start = 0; start <= n - len; start++) {
int end = start + len - 1;
dp[start][end] = dp[start][end - 1] * 10 + s[end] - '0';
sum += dp[start][end];
}
}
return sum;
}
int main()
{
int t, n, item;
cout << "enter the string: ";
string s;
cin >> s;
cout << "sum of all possible substring is: " << my(s, s.length()) << endl;
return 0;
}
Output:
输出：
RUN 1:
enter the string: 17678
sum of all possible substring is: 29011
RUN 2:
enter the string: 326
sum of all possible substring is: 395
翻译自: https://www.includehelp.com/icp/sum-of-all-substrings-of-a-number.aspx
所有子序列的逆序对总和
