所有子序列的逆序對總和

Problem statement:

問題陳述：

Given an integer, S represented as a string, get the sum of all possible substrings of this string.

給定一個以字符串形式表示的整數S ，得到該字符串所有可能的子字符串的和 。

Input:

輸入：

A string S that representing the number.

代表數字的字符串S。

Output:

輸出：

Print sum of all possible substrings as required result.

根據要求的結果打印所有可能的子字符串的總和。

Constraints:

限制條件：

1 <= T <= 100 1 <= S <= 10¹²

Example:

例：

Input:1234 326Output: 1670 395

Explanation:

說明：

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:

解決方法：

The solution approach is by storing the substring sums to compute the exact next substring sum

解決方法是通過存儲子字符串和以計算確切的下一個子字符串和

Create dp[n][n] to store substring sums;

創建dp [n] [n]來存儲子字符串和；

Initialize sum=0 which will be our final result;

初始化sum = 0，這將是我們的最終結果；

Base case computation (single length substrings),

基本案例計算(單長度子字符串)，

for i=0 to n-1,n= string lengthdp[i][i]=s[i] -'0'; //s[i]-'0' gives the digit actuallysum+=dp[i][i]; end for

Till now we have computed all single digit substrings,

到現在為止，我們已經計算了所有個位數的子字符串，

for substring length,len=2 to nfor 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 for

Sum is the final result.

總和是最終結果。

All the statements are self-explanatory except the one which is the fundamental idea of the entire storing process. That is the below one,

所有陳述都是不言自明的，只是整個存儲過程的基本思想。 那是下一個，

dp[start][end]=dp[start][end-1]*10+s[end]-'0';

Let's check this with an example,

我們來看一個例子，

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

讓我們開始同樣的例子

N=4 here

N = 4這里

So, we need to fill up a 4X4 DP table,

因此，我們需要填寫4X4 DP表，

After filling the base case,

裝完基本外殼后，

Now, I am computing for len=2

現在，我正在計算len = 2

Start=0, end=1

開始= 0，結束= 1

Start=1, end=2

開始= 1，結束= 2

Start=2, end=3

開始= 2，結束= 3

For len =3

對于len = 3

Start=0, end=2

開始= 0，結束= 2

Start=1, end=3

開始= 1，結束= 3

Len=4

Len = 4

Start=0, end=3

開始= 0，結束= 3

At each step we have summed up, so result is stored at sum.

在每一步我們都進行了總結，因此結果被存儲在總和中。

C++ Implementation:

C ++實現：

#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:

輸出：

RUN 1: enter the string: 17678 sum of all possible substring is: 29011RUN 2: enter the string: 326 sum of all possible substring is: 395

翻譯自: https://www.includehelp.com/icp/sum-of-all-substrings-of-a-number.aspx

所有子序列的逆序對總和

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