📌  相关文章
📜  最多 4 个 X 次、最多 5 个 Y 次和最多 6 个 Z 次形成的所有数字的总和

📅  最后修改于: 2021-09-17 06:59:03             🧑  作者: Mango

给定三个整数X 、 Y和Z ，任务是在模 10^9+7 下找到最多4个X次、最多5个Y次和最多6个Z次形成的所有数字的总和。
例子： 

Input: X = 1, Y = 1, Z = 1 
Output: 3675
Explanation:
4 + 5 + 6 + 45 + 54 + 56 
+ 65 + 46 + 64 + 456 + 465 
+ 546 + 564 + 645 + 654 = 3675

Input: X = 4, Y = 5, Z = 6
Output: 129422134

方法：

  • 由于这个问题具有子问题重叠和最优子结构的性质，因此可以使用动态规划来解决它。
  • 对于所有i < x, j < y, j < z ，需要具有精确i 4s 、 j 5s和k 6s的数字才能获得所需的总和。
  • 因此，dp 数组exactnum[i][j][k]将存储具有精确i 4s 、 j 5s和k 6s的数字的精确计数。
  • 如果exactnum[i – 1][j][k] 、 exactnum[i][j – 1][k]和exactnum[i][j][k – 1]是已知的，那么可以观察到这些的总和是必需的答案，除非exactnum[i – 1][j][k] 、 exactnum[i][j – 1][k]或exactnum[i][j][k – 1 ]不存在。在这种情况下，只需跳过它。
  • Exactsum[i][j][k]以与以下相同的方式存储具有i 个4、 j 个5 和k 个6 的精确数的总和
exactsum[i][j][k] = 10 * (exactsum[i - 1][j][k] 
                        + exactsum[i][j - 1][k] 
                        + exactsum[i][j][k - 1]) 
                  + 4 * exactnum[i - 1][j][k] 
                  + 5 * exactnum[i][j - 1][k] 
                  + 6 * exactnum[i][j][k - 1]

下面是上述方法的实现：

C++
// C++ program to find sum of all numbers
// formed having 4 atmost X times, 5 atmost
// Y times and 6 atmost Z times
#include 
using namespace std;
 
const int N = 101;
const int mod = 1e9 + 7;
 
// exactsum[i][j][k] stores the sum of
// all the numbers having exact
// i 4's, j 5's and k 6's
int exactsum[N][N][N];
 
// exactnum[i][j][k] stores numbers
// of numbers having exact
// i 4's, j 5's and k 6's
int exactnum[N][N][N];
 
// Utility function to calculate the
// sum for x 4's, y 5's and z 6's
int getSum(int x, int y, int z)
{
    int ans = 0;
    exactnum[0][0][0] = 1;
    for (int i = 0; i <= x; ++i) {
        for (int j = 0; j <= y; ++j) {
            for (int k = 0; k <= z; ++k) {
 
                // Computing exactsum[i][j][k]
                // as explained above
                if (i > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i - 1][j][k] * 10
                            + 4 * exactnum[i - 1][j][k])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i - 1][j][k] % mod;
                }
                if (j > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i][j - 1][k] * 10
                            + 5 * exactnum[i][j - 1][k])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i][j - 1][k] % mod;
                }
                if (k > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i][j][k - 1] * 10
                            + 6 * exactnum[i][j][k - 1])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i][j][k - 1] % mod;
                }
 
                ans += exactsum[i][j][k] % mod;
                ans %= mod;
            }
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int x = 1, y = 1, z = 1;
 
    cout << (getSum(x, y, z) % mod);
 
    return 0;
}

Java
// Java program to find sum of all numbers
// formed having 4 atmost X times, 5 atmost
// Y times and 6 atmost Z times
     
class GFG
{
     
    static int N = 101;
    static int mod = (int)1e9 + 7;
     
    // exactsum[i][j][k] stores the sum of
    // all the numbers having exact
    // i 4's, j 5's and k 6's
    static int exactsum[][][] = new int[N][N][N];
     
    // exactnum[i][j][k] stores numbers
    // of numbers having exact
    // i 4's, j 5's and k 6's
    static int exactnum[][][] = new int[N][N][N];
     
    // Utility function to calculate the
    // sum for x 4's, y 5's and z 6's
    static int getSum(int x, int y, int z)
    {
        int ans = 0;
        exactnum[0][0][0] = 1;
        for (int i = 0; i <= x; ++i)
        {
            for (int j = 0; j <= y; ++j)
            {
                for (int k = 0; k <= z; ++k)
                {
     
                    // Computing exactsum[i][j][k]
                    // as explained above
                    if (i > 0)
                    {
                        exactsum[i][j][k]
                        += (exactsum[i - 1][j][k] * 10
                        + 4 * exactnum[i - 1][j][k]) % mod;
                         
                        exactnum[i][j][k]
                        += exactnum[i - 1][j][k] % mod;
                    }
                    if (j > 0)
                    {
                        exactsum[i][j][k]
                        += (exactsum[i][j - 1][k] * 10
                        + 5 * exactnum[i][j - 1][k]) % mod;
                         
                        exactnum[i][j][k]
                        += exactnum[i][j - 1][k] % mod;
                    }
                    if (k > 0)
                    {
                        exactsum[i][j][k]
                        += (exactsum[i][j][k - 1] * 10
                        + 6 * exactnum[i][j][k - 1]) % mod;
                         
                        exactnum[i][j][k]
                        += exactnum[i][j][k - 1] % mod;
                    }
     
                    ans += exactsum[i][j][k] % mod;
                    ans %= mod;
                }
            }
        }
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int x = 1, y = 1, z = 1;
     
        System.out.println(getSum(x, y, z) % mod);
     
    }
}
 
// This code is contributed by AnkitRai01

Python3
# Python3 program to find sum of all numbers
# formed having 4 atmost X times, 5 atmost
# Y times and 6 atmost Z times
import numpy as np
 
N = 101;
mod = int(1e9) + 7;
 
# exactsum[i][j][k] stores the sum of
# all the numbers having exact
# i 4's, j 5's and k 6's
exactsum = np.zeros((N, N, N));
 
# exactnum[i][j][k] stores numbers
# of numbers having exact
# i 4's, j 5's and k 6's
exactnum = np.zeros((N, N, N));
 
# Utility function to calculate the
# sum for x 4's, y 5's and z 6's
def getSum(x, y, z) :
    ans = 0;
    exactnum[0][0][0] = 1;
    for i in range(x + 1) :
        for j in range(y + 1) :
            for k in range(z + 1) :
 
                # Computing exactsum[i][j][k]
                # as explained above
                if (i > 0) :
                    exactsum[i][j][k] += (exactsum[i - 1][j][k] * 10 +
                                            4 * exactnum[i - 1][j][k]) % mod;
                                             
                    exactnum[i][j][k] += exactnum[i - 1][j][k] % mod;
                 
                if (j > 0) :
                    exactsum[i][j][k] += (exactsum[i][j - 1][k] * 10+
                                        5 * exactnum[i][j - 1][k]) % mod;
                                         
                    exactnum[i][j][k] += exactnum[i][j - 1][k] % mod;
                 
                if (k > 0) :
                    exactsum[i][j][k] += (exactsum[i][j][k - 1] * 10
                                            + 6 * exactnum[i][j][k - 1]) % mod;
                    exactnum[i][j][k] += exactnum[i][j][k - 1] % mod;
 
                ans += exactsum[i][j][k] % mod;
                ans %= mod;
                 
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    x = 1; y = 1; z = 1;
 
    print((getSum(x, y, z) % mod));
 
# This code is contributed by AnkitRai01

C#
// C# program to find sum of all numbers
// formed having 4 atmost X times, 5 atmost
// Y times and 6 atmost Z times
using System;
 
class GFG
{
     
    static int N = 101;
    static int mod = (int)1e9 + 7;
     
    // exactsum[i][j][k] stores the sum of
    // all the numbers having exact
    // i 4's, j 5's and k 6's
    static int [,,]exactsum = new int[N, N, N];
     
    // exactnum[i][j][k] stores numbers
    // of numbers having exact
    // i 4's, j 5's and k 6's
    static int [,,]exactnum= new int[N, N, N];
     
    // Utility function to calculate the
    // sum for x 4's, y 5's and z 6's
    static int getSum(int x, int y, int z)
    {
        int ans = 0;
        exactnum[0, 0, 0] = 1;
        for (int i = 0; i <= x; ++i)
        {
            for (int j = 0; j <= y; ++j)
            {
                for (int k = 0; k <= z; ++k)
                {
     
                    // Computing exactsum[i, j, k]
                    // as explained above
                    if (i > 0)
                    {
                        exactsum[i, j, k]
                        += (exactsum[i - 1, j, k] * 10
                        + 4 * exactnum[i - 1, j, k]) % mod;
                         
                        exactnum[i, j, k]
                        += exactnum[i - 1, j, k] % mod;
                    }
                    if (j > 0)
                    {
                        exactsum[i, j, k]
                        += (exactsum[i, j - 1, k] * 10
                        + 5 * exactnum[i, j - 1, k]) % mod;
                         
                        exactnum[i, j, k]
                        += exactnum[i, j - 1, k] % mod;
                    }
                    if (k > 0)
                    {
                        exactsum[i, j, k]
                        += (exactsum[i, j, k - 1] * 10
                        + 6 * exactnum[i, j, k - 1]) % mod;
                         
                        exactnum[i, j, k]
                        += exactnum[i, j, k - 1] % mod;
                    }
     
                    ans += exactsum[i, j, k] % mod;
                    ans %= mod;
                }
            }
        }
        return ans;
    }
     
    // Driver code
    public static void Main ()
    {
        int x = 1, y = 1, z = 1;
     
        Console.WriteLine(getSum(x, y, z) % mod);
     
    }
}
     
// This code is contributed by AnkitRai01

Javascript

输出： 
3675

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。