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📜  通过替换？获得的整数计数在给定的字符串，除以13所得的余数为5

📅  最后修改于: 2021-04-22 08:55:22             🧑  作者: Mango

给定长度为N的字符串str 。任务是找到通过替换‘?’获得的整数数量。具有任何数字，以使形成的整数在除以13时得到余数5 。
数字也可以从零开始。答案可能非常大，因此，以10 9 + 7为模输出答案。
例子：

方法：假设dp [i] [j]是创建与给定模式的前i位一致并与j模13相等的i位数字的方式的数量。作为我们的基本情况，对于从1到12的i ， dp [0] [i] = 0 ，并且dp [0] [0] = 1 (因为长度为零的数字的值为零，因此mod 13的值为零)。
请注意，将数字k附加到j mod 13的数字的末尾可得到与10j + k mod 13一致的数字。我们使用这个事实来执行我们的过渡。对于每个状态， i dp [i] [j]迭代k的可能值。 (如果s [i] =’?’ ，则k会有十个选择，否则，只有一个选择。)然后，将dp [i] [j]加到dp [i + 1] [( 10j + k)％13]。
为了得到最终答案，我们可以简单地打印dp [N] [5] 。
下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
#define MOD (int)(1e9 + 7)
 
// Function to find the count of integers
// obtained by replacing '?' in a given
// string such that formed integer
// gives remainder 5 when it is divided by 13
int modulo_13(string s, int n)
{
    long long dp[n + 1][13] = { { 0 } };
 
    // Initialise
    dp[0][0] = 1;
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < 10; j++) {
            int nxt = s[i] - '0';
 
            // Place digit j at ? position
            if (s[i] == '?')
                nxt = j;
 
            // Get the remainder
            for (int k = 0; k < 13; k++) {
                int rem = (10 * k + nxt) % 13;
                dp[i + 1][rem] += dp[i][k];
                dp[i + 1][rem] %= MOD;
            }
            if (s[i] != '?')
                break;
        }
    }
 
    // Return the required answer
    return (int)dp[n][5];
}
 
// Driver code
int main()
{
    string s = "?44";
    int n = s.size();
 
    cout << modulo_13(s, n);
 
    return 0;
}

Java
// Java implementation of the approach
class GFG
{
static int MOD = (int)(1e9 + 7);
 
// Function to find the count of integers
// obtained by replacing '?' in a given
// string such that formed integer
// gives remainder 5 when it is divided by 13
static int modulo_13(String s, int n)
{
    long [][]dp = new long[n + 1][13];
 
    // Initialise
    dp[0][0] = 1;
 
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 10; j++)
        {
            int nxt = s.charAt(i) - '0';
 
            // Place digit j at ? position
            if (s.charAt(i) == '?')
                nxt = j;
 
            // Get the remainder
            for (int k = 0; k < 13; k++)
            {
                int rem = (10 * k + nxt) % 13;
                dp[i + 1][rem] += dp[i][k];
                dp[i + 1][rem] %= MOD;
            }
            if (s.charAt(i) != '?')
                break;
        }
    }
 
    // Return the required answer
    return (int)dp[n][5];
}
 
// Driver code
public static void main(String []args)
{
    String s = "?44";
    int n = s.length();
 
    System.out.println(modulo_13(s, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3
# Python3 implementation of the approach
import numpy as np
 
MOD = (int)(1e9 + 7)
 
# Function to find the count of integers
# obtained by replacing '?' in a given
# string such that formed integer
# gives remainder 5 when it is divided by 13
def modulo_13(s, n) :
     
    dp = np.zeros((n + 1, 13));
     
    # Initialise
    dp[0][0] = 1;
 
    for i in range(n) :
        for j in range(10) :
            nxt = ord(s[i]) - ord('0');
 
            # Place digit j at ? position
            if (s[i] == '?') :
                nxt = j;
 
            # Get the remainder
            for k in range(13) :
                rem = (10 * k + nxt) % 13;
                dp[i + 1][rem] += dp[i][k];
                dp[i + 1][rem] %= MOD;
         
            if (s[i] != '?') :
                break;
 
    # Return the required answer
    return int(dp[n][5]);
 
# Driver code
if __name__ == "__main__" :
    s = "?44";
    n = len(s);
 
    print(modulo_13(s, n));
 
# This code is contributed by AnkitRai01

C#
// C# implementation of the approach
using System;
                     
class GFG
{
 
static int MOD = (int)(1e9 + 7);
 
// Function to find the count of integers
// obtained by replacing '?' in a given
// string such that formed integer
// gives remainder 5 when it is divided by 13
static int modulo_13(String s, int n)
{
    long [,]dp = new long[n + 1, 13];
 
    // Initialise
    dp[0, 0] = 1;
 
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 10; j++)
        {
            int nxt = s[i] - '0';
 
            // Place digit j at ? position
            if (s[i] == '?')
                nxt = j;
 
            // Get the remainder
            for (int k = 0; k < 13; k++)
            {
                int rem = (10 * k + nxt) % 13;
                dp[i + 1, rem] += dp[i, k];
                dp[i + 1, rem] %= MOD;
            }
            if (s[i] != '?')
                break;
        }
    }
 
    // Return the required answer
    return (int)dp[n,5];
}
 
// Driver code
public static void Main(String []args)
{
    String s = "?44";
    int n = s.Length;
 
    Console.WriteLine(modulo_13(s, n));
}
}
 
// This code is contributed by 29AjayKumar

输出： 
1

时间复杂度： O(100 * N)

辅助空间： O(100 * N)