編程 小數位數

Problem statement:

問題陳述：

Given the number of digits n, find the count of total non-decreasing numbers with n digits.

給定位數n ，找到具有n位數字的非遞減總數。

A number is non-decreasing if every digit (except the first one) is greater than or equal to the previous digit. For example, 22,223, 45567, 899, are non-decreasing numbers whereas 321 or 322 are not.

如果每個數字(第一個數字除外)都大于或等于前一個數字，則數字不減。 例如，22,223、45567、899是不減數字，而321或322不是。

Input:

輸入：

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains the integer n.

輸入的第一行包含一個整數T，表示測試用例的數量。 然后是T測試用例。 每個測試用例的第一行包含整數n 。

Output:

輸出：

Print the count of total non-decreasing numbers with n digits for each test case in a new line. You may need to use unsigned long long int as the count can be very large.

在新行中為每個測試用例用n位數字打印非降序總數。 您可能需要使用unsigned long long int，因為計數可能非常大。

Constraints:

限制條件：

1 <= T <= 100
1 <= n <= 200

Example:

例：

Input:
No of test cases, 3
n=1
n=2
n=3
Output:
For n=1
Total count is: 10
For n=2
Total count is: 55
For n=3
Total count is: 220

Explanation:

說明：

For n=1,
The non-decreasing numbers are basically 0 to 9, 
counting up to 10
For, n=2,
The non-decreasing numbers can be
00
01
02
..
11
12
13
14
15
.. so on total 55

Solution Approach:

解決方法：

The solution can be recursive. In recursion our strategy will be to build the string (read: number) such that at each recursive call the number to be appended would be necessarily bigger (or equal) than(to) the last one.

解決方案可以是遞歸的。 在遞歸中，我們的策略是構建字符串(讀取：number)，以便在每次遞歸調用時，要附加的數字必然大于(或等于)最后一個數字。

Say, at any recursion call,

說，在任何遞歸調用中，

The number already constructed is x₁x₂...x_i where i<n, So at the recursive call we are allowed to append digits only which are equal to greater to x_i.

已經構造的數字是x ₁ x ₂ ... x _i ，其中i <n ，因此在遞歸調用中，我們只允許附加等于x _i的數字 。

So, let's formulate the recursion

因此，讓我們制定遞歸

Say, the recursive function is computerecur(int index, int last, int n)

說，遞歸函數是computerecur(int index，int last，int n)

Where,

哪里，

• index = current position to append digit

  索引 =當前位置追加數字

• Last = the previous digit

  最后 =前一位

• n = number of total digits

  n =總位數

unsigned long long int computerecur (int index,int last,int n){
// base case
if(index has reached the last digit)
return 1;
unsigned long long int sum=0;
for digit to append at current index,i=0 to 9
if(i>=last)
// recur if I can be appended
sum + =computerecur(index+1,i,n); 
end for
return sum    
End function

So the basic idea to the recursion is that if it's a valid digit to append (depending on the last digit) then append it and recur for the remaining digits.

因此，遞歸的基本思想是，如果要追加的有效數字(取決于最后一位)，則將其追加并針對剩余的數字進行遞歸。

Now the call from the main() function should be done by appending the first digit already (basically we will keep that first digit as last digit position for the recursion call).

現在，應該通過已經附加了第一個數字來完成對main()函數的調用(基本上，我們將把該第一個數字保留為遞歸調用的最后一個數字位置)。

So at main,

所以總的來說

We will call as,

我們稱之為

unsigned long long int result=0;
for i=0 to 9 //starting digits
// index=0, starting digit assigned as 
// last digit for recursion
result+=computerecur(0,i,n); 
end for

The result is the ultimate result.

結果就是最終結果。

I would suggest you draw the recursion tree to have a better understanding, Take n=3 and do yourself.

我建議您繪制遞歸樹以更好地理解，取n = 3并自己做。

For, n=2, I will brief the tree below

對于n = 2 ，我將簡要介紹下面的樹

For starting digit 0
Computerecur(0,0,2) 
Index!=n-1
So
Goes to the loop
And then 
It calls to
Computerecur(1,0,2) // it's for number 00
Computerecur(1,1,2) // it's for number 01
Computerecur(1,2,2) // it's for number 02
Computerecur(1,3,2) // it's for number 03
Computerecur(1,4,2) // it's for number 04
Computerecur(1,5,2) // it's for number 05
Computerecur(1,6,2) // it's for number 06
Computerecur(1,7,2) // it's for number 07
Computerecur(1,8,2) // it's for number 08
Computerecur(1,9,2) // it's for number 09
So on
...

Now, it's pretty easy to infer that it leads to many overlapping sub-problem and hence we need dynamic programming to store the results of overlapping sub-problems. That's why I have used the memoization technique to store the already computed sub-problem results. See the below implementation to understand the memorization part.

現在，很容易推斷出它會導致許多子問題重疊，因此我們需要動態編程來存儲子問題重疊的結果。 這就是為什么我使用記憶技術來存儲已經計算出的子問題結果的原因。 請參閱以下實現以了解記憶部分。

C++ Implementation:

C ++實現：

#include <bits/stdc++.h>
using namespace std;
unsigned long long int dp[501][10];
unsigned long long int my(int index, int last, int n)
{
if (index == n - 1)
return 1;
// memorization, don't compute again what is already computed
if (dp[index][last] != -1) 
return dp[index][last];
unsigned long long int sum = 0;
for (int i = 0; i <= 9; i++) {
if (i >= last)
sum += my(index + 1, i, n);
}
dp[index][last] = sum;
return dp[index][last];
}
unsigned long long int compute(int n)
{
unsigned long long int sum = 0;
for (int i = 0; i <= 9; i++) {
sum += my(0, i, n);
}
return sum;
}
int main()
{
int t, n, item;
cout << "enter number of testcase\n";
scanf("%d", &t);
for (int i = 0; i < t; i++) {
cout << "Enter the number of digits,n:\n";
scanf("%d", &n);
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= 9; j++) {
dp[i][j] = -1;
}
}
cout << "number of non-decreasing number with " << n;
cout << " digits are: " << compute(n) << endl;
}
return 0;
}

Output:

輸出：

enter number of testcase
3
Enter the number of digits,n:
2
number of non-decreasing number with 2 digits are: 55
Enter the number of digits,n:
3
number of non-decreasing number with 3 digits are: 220
Enter the number of digits,n:
6
number of non-decreasing number with 6 digits are: 5005

翻譯自: https://www.includehelp.com/icp/total-number-of-non-decreasing-numbers-with-n-digits-using-dynamic-programming.aspx

編程 小數位數