📌  相关文章
📜  数字d恰好出现K次的范围内的数字计数

📅  最后修改于: 2021-06-25 11:27:16             🧑  作者: Mango

给定两个代表范围的正整数L和R以及另外两个正整数d和K。任务是找到数字d恰好出现K次的范围内的数字计数。

例子：

先决条件：Digit DP

方法：首先，如果我们能够计算所需的数字，直到R，即在[0，R]范围内，则可以通过求解从零到R的值，然后减去该值，就可以轻松地在[L，R]范围内得出答案。从零到L – 1求解后得到的答案。现在，我们需要定义DP状态。
DP州：

  • 既然我们可以考虑我们的号码作为数字序列，一个状态是，我们目前的位置，这个位置可以有值从0到18，如果我们处理的是数字高达10 18。在每个递归调用中，我们尝试通过从0到9放置一个数字来从左到右构建序列。
  • 第二种状态是计数，它定义了次数，到目前为止，我们已经将数字d放置了。
  • 另一个状态是布尔变量tight ，它指示我们要构建的数字已经小于R，因此在接下来的递归调用中，我们可以将0到9之间的任何数字放置。如果数字没有变小，则最大限制为我们可以放置的数字是R中当前位置的数字。
  • 最后一个状态也是布尔变量nonz ，如果正在构建的数字中有任何前导零，则有助于考虑这种情况，我们不需要对它们进行计数。

    在最后的递归调用中，如果我们在最后一个位置，如果数字d的计数等于K，则返回1，否则返回0。

    下面是上述方法的实现：

    C++
    // CPP Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
#include 
using namespace std;
  
const int M = 20;
  
// states - position, count, tight, nonz
int dp[M][M][2][2];
  
// d is required digit and K is occurrence
int d, K;
  
// This function returns the count of
// required numbers from 0 to num
int count(int pos, int cnt, int tight,
          int nonz, vector num)
{
    // Last position
    if (pos == num.size()) {
        if (cnt == K)
            return 1;
        return 0;
    }
  
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight][nonz] != -1)
        return dp[pos][cnt][tight][nonz];
  
    int ans = 0;
  
    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = (tight ? 9 : num[pos]);
  
    for (int dig = 0; dig <= limit; dig++) {
        int currCnt = cnt;
  
        // Nonz is true if we placed a non
        // zero digit at the starting of
        // the number
        if (dig == d) {
            if (d != 0 || (!d && nonz))
                currCnt++;
        }
  
        int currTight = tight;
  
        // At this position, number becomes
        // smaller
        if (dig < num[pos])
            currTight = 1;
  
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        ans += count(pos + 1, currCnt,
                     currTight, nonz || (dig != 0), num);
    }
    return dp[pos][cnt][tight][nonz] = ans;
}
  
// Function to convert x into its digit vector and uses
// count() function to return the required count
int solve(int x)
{
    vector num;
    while (x) {
        num.push_back(x % 10);
        x /= 10;
    }
    reverse(num.begin(), num.end());
  
    // Initialize dp
    memset(dp, -1, sizeof(dp));
    return count(0, 0, 0, 0, num);
}
  
// Driver Code to test above functions
int main()
{
    int L = 11, R = 100;
    d = 2, K = 1;
    cout << solve(R) - solve(L - 1) << endl;
  
    return 0;
}

    Java
    // Java Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
import java.util.*;
class Solution
{
static final int M = 20;
   
// states - position, count, tight, nonz
static int dp[][][][]= new int[M][M][2][2];
   
// d is required digit and K is occurrence
static int d, K;
   
// This function returns the count of
// required numbers from 0 to num
static int count(int pos, int cnt, int tight,
          int nonz, Vector num)
{
    // Last position
    if (pos == num.size()) {
        if (cnt == K)
            return 1;
        return 0;
    }
   
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight][nonz] != -1)
        return dp[pos][cnt][tight][nonz];
   
    int ans = 0;
   
    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = ((tight !=0)? 9 : num.get(pos));
   
    for (int dig = 0; dig <= limit; dig++) {
        int currCnt = cnt;
   
        // Nonz is true if we placed a non
        // zero digit at the starting of
        // the number
        if (dig == d) {
            if (d != 0 || (d==0 && nonz!=0))
                currCnt++;
        }
   
        int currTight = tight;
   
        // At this position, number becomes
        // smaller
        if (dig < num.get(pos))
            currTight = 1;
   
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        ans += count(pos + 1, currCnt,
                     currTight, (dig != 0?1:0), num);
    }
    return dp[pos][cnt][tight][nonz] = ans;
}
   
// Function to convert x into its digit vector and uses
// count() function to return the required count
static int solve(int x)
{
    Vector num= new Vector();
    while (x!=0) {
        num.add(x % 10);
        x /= 10;
    }
    Collections.reverse(num);
   
    // Initialize dp
    for(int i=0;i

    Python3
    # Python Program to find the count of
# numbers in a range where digit d
# occurs exactly K times
M = 20
  
# states - position, count, tight, nonz
dp = []
  
# d is required digit and K is occurrence
d, K = None, None
  
# This function returns the count of
# required numbers from 0 to num
def count(pos, cnt, tight, nonz, num: list):
  
    # Last position
    if pos == len(num):
        if cnt == K:
            return 1
        return 0
  
    # If this result is already computed
    # simply return it
    if dp[pos][cnt][tight][nonz] != -1:
        return dp[pos][cnt][tight][nonz]
  
    ans = 0
  
    # Maximum limit upto which we can place
    # digit. If tight is 1, means number has
    # already become smaller so we can place
    # any digit, otherwise num[pos]
    limit = 9 if tight else num[pos]
  
    for dig in range(limit + 1):
        currCnt = cnt
  
        # Nonz is true if we placed a non
        # zero digit at the starting of
        # the number
        if dig == d:
            if d != 0 or not d and nonz:
                currCnt += 1
  
        currTight = tight
  
        # At this position, number becomes
        # smaller
        if dig < num[pos]:
            currTight = 1
  
        # Next recursive call, also set nonz
        # to 1 if current digit is non zero
        ans += count(pos + 1, currCnt, 
                currTight, (nonz or dig != 0), num)
  
    dp[pos][cnt][tight][nonz] = ans
    return dp[pos][cnt][tight][nonz]
  
  
# Function to convert x into its digit vector and uses
# count() function to return the required count
def solve(x):
    global dp, K, d
  
    num = []
    while x:
        num.append(x % 10)
        x //= 10
  
    num.reverse()
  
    # Initialize dp
    dp = [[[[-1, -1] for i in range(2)] 
            for j in range(M)] for k in range(M)]
    return count(0, 0, 0, 0, num)
  
# Driver Code
if __name__ == "__main__":
  
    L = 11
    R = 100
    d = 2
    K = 1
    print(solve(R) - solve(L - 1))
  
# This code is contributed by
# sanjeev2552

    C#
    // C# Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
using System;
using System.Collections.Generic; 
  
class GFG
{
    static readonly int M = 20;
  
    // states - position, count, tight, nonz
    static int [,,,]dp= new int[M, M, 2, 2];
  
    // d is required digit and K is occurrence
    static int d, K;
  
    // This function returns the count of
    // required numbers from 0 to num
    static int count(int pos, int cnt, int tight,
            int nonz, List num)
    {
        // Last position
        if (pos == num.Count) 
        {
            if (cnt == K)
                return 1;
            return 0;
        }
  
        // If this result is already computed
        // simply return it
        if (dp[pos, cnt, tight, nonz] != -1)
            return dp[pos, cnt, tight, nonz];
  
        int ans = 0;
  
        // Maximum limit upto which we can place
        // digit. If tight is 1, means number has
        // already become smaller so we can place
        // any digit, otherwise num[pos]
        int limit = ((tight != 0) ? 9 : num[pos]);
  
        for (int dig = 0; dig <= limit; dig++)
        {
            int currCnt = cnt;
  
            // Nonz is true if we placed a non
            // zero digit at the starting of
            // the number
            if (dig == d) 
            {
                if (d != 0 || (d == 0 && nonz != 0))
                    currCnt++;
            }
  
            int currTight = tight;
  
            // At this position, number becomes
            // smaller
            if (dig < num[pos])
                currTight = 1;
  
            // Next recursive call, also set nonz
            // to 1 if current digit is non zero
            ans += count(pos + 1, currCnt,
                        currTight, (dig != 0 ? 1 : 0), num);
        }
        return dp[pos, cnt, tight, nonz] = ans;
    }
  
    // Function to convert x into its 
    // digit vector and uses count() 
    // function to return the required count
    static int solve(int x)
    {
        List num = new List();
        while (x != 0) 
        {
            num.Add(x % 10);
            x /= 10;
        }
        num.Reverse();
  
        // Initialize dp
        for(int i = 0; i < M; i++)
            for(int j = 0; j < M; j++)
                for(int k = 0; k < 2; k++)
                    for(int l = 0; l < 2; l++)
                        dp[i, j, k, l]=-1;
  
        return count(0, 0, 0, 0, num);
    }
  
    // Driver Code 
    public static void Main()
    {
        int L = 11, R = 100;
        d = 2; K = 1;
        Console.Write( solve(R) - solve(L - 1) );
    }
}
  
// This code is contributed by Rajput-JI

    输出： 
    17

    如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。