vb 導出整數(shù) 科學計數(shù)法

Problem statement:

問題陳述：

Given two positive integer n and m, find how many arrays of size n that can be formed such that:

給定兩個正整數(shù)n和m ，找出可以形成多少個大小為n的數(shù)組：

Each element of the array is in the range [1, m]

數(shù)組的每個元素都在[1，m]范圍內(nèi)

Any adjacent element pair is divisible, i.e., that one of them divides another. Either element A[i] divides A[i + 1] or A[i + 1] divides A[i].

任何相鄰的元素對都是可分割的 ，即，其中一個元素對另一個元素 。 元素A [i]除以A [i + 1]或A [i + 1]除以A [i] 。

Input:

輸入：

Only one line with two integer, n & m respectively.

只有一行包含兩個整數(shù)，分別為n和m 。

Output:

輸出：

Print number of different possible ways to create the array. Since the output could be long take modulo 10^9+7.

打印創(chuàng)建數(shù)組的各種可能方式的數(shù)量。 由于輸出可能很長，取模10 ^ 9 + 7 。

Constraints:

限制條件：

1<=n, m<=100

Example:

例：

Input: n = 3, m = 2.Output: 8Explanation:{1,1,1},{1, 1, 2}, {1, 2, 1}, {1, 2, 2}, {2, 1, 1},{2,1,2}, {2,2,1}, {2,2,2} are possible arrays.Input: n = 1, m = 5.Output: 5Explanation:{1}, {2}, {3}, {4}, {5}

Solution Approach:

解決方法：

The above problem is a great example of recursion. What can be the recursive function and how we can formulate.

上面的問題是遞歸的一個很好的例子。 什么是遞歸函數(shù)，以及我們?nèi)绾喂交?

Say,

說，

Let

讓

F(n, m)?= number of ways for array size n and range 1 to m

F(n，m) =數(shù)組大小為n且范圍為1到m的路徑數(shù)

Now,

現(xiàn)在，

We actually can try picking every element from raging 1 to m and try recurring for other elements

實際上，我們可以嘗試從1到m范圍內(nèi)選取每個元素，然后嘗試對其他元素進行重復

So, the function can be written like:

因此，該函數(shù)可以這樣寫：

Function: NumberofWays(cur_index,lastelement,n,m)

So, to describe the arguments,

因此，為了描述這些論點，

cur_index is your current index and the last element is the previous index value assigned. So basically need to check which value with range 1 to m fits in the cur_index such that the divisibility constraint satisfies.

cur_index是當前索引，最后一個元素是分配的前一個索引值。 因此，基本上需要檢查范圍在1到m之間的哪個值適合cur_index ，以便除數(shù)約束滿足。

So, to describe the body of the function

因此，要描述功能的主體

Function NumberofWays(cur_index,lastelement,n,m) // formed the array completelyif(cur_index==n)return 1;sum=0// any element in the rangefor j=1 to m // if divisible then lastelement,// j can be adjacent pairif(j%lastelement==0 || lastelement%j==0)// recur for rest of the elmentssum=(sum%MOD+ NumberofWays(cur_index+1,j,n,m)%MOD)%MOD; end ifend for End function

Now the above recursive function generates many overlapping sub-problem and that's why we use the top-down DP approach to store already computed sub-problem results.
Below is the implementation with adding memoization.

現(xiàn)在，上面的遞歸函數(shù)會生成許多重疊的子問題，這就是為什么我們使用自上而下的DP方法來存儲已經(jīng)計算出的子問題結(jié)果的原因。
下面是添加備忘錄的實現(xiàn)。

Initiate a 2D DP array with -1

使用-1啟動2D DP陣列

Function NumberofWays(cur_index,lastelement,n,m)// formed the array completelyif(cur_index==n)return 1;// if solution to sub problem already exitsif(dp[cur_index][lastelement]!=-1) return dpdp[cur_index][lastelement]; sum=0for j=1 to m // any element in the range// if divisible then lastelement,j can be adjacent pairif(j%lastelement==0 || lastelement%j==0)// recur for rest of the elmentssum=(sum%MOD+ NumberofWays(cur_index+1,j,n,m)%MOD)%MOD; end ifend forDp[cur_index][lastelement]=sumReturn Dp[cur_index][lastelement] End function

C++ Implementation:

C ++實現(xiàn)：

#include <bits/stdc++.h> using namespace std;#define MOD 1000000007int dp[101][101];int countarray(int index, int i, int n, int m) {// if solution to sub problem already exitsif (dp[index][i] != -1)return dp[index][i];if (index == n)return 1;int sum = 0;//any element in the rangefor (int j = 1; j <= m; j++) {// if divisible then i,j can be adjacent pairif (j % i == 0 || i % j == 0) {// recur for rest of the elmentssum = (sum % MOD + countarray(index + 1, j, n, m) % MOD) % MOD;}}dp[index][i] = sum;return dp[index][i]; }int main() {int n, m;cout << "Enter value of n:\n";cin >> n;cout << "Enter value of m:\n";cin >> m;// initialize DP matrixfor (int i = 0; i <= n; i++) {for (int j = 0; j <= m; j++) {dp[i][j] = -1;}}cout << "number of ways are: " << countarray(0, 1, n, m) << endl;return 0; }

Output:

輸出：

Enter value of n: 3 Enter value of m: 2 number of ways are: 8

翻譯自: https://www.includehelp.com/icp/count-of-divisible-array.aspx

vb 導出整數(shù) 科學計數(shù)法

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