通过将每个数字替换为 X 的乘积，检查是否可以在 M 步中将 S 转换为 Target

给定两个数字字符串S和Target，以及整数X, M 。任务是检查是否可以使用以下操作将字符串S转换为字符串Target ：

• 在一个操作中，需要将S中的每个数字替换为其与X 的乘积，即S[i] = (S[I]*X)。
• 上述操作必须准确地进行M次。

例子：

Input: S = “1234”, Target = “2550525100”, X = 5, M = 2
Output: YES
Explanation:
Initially S=1234
After the 1st, level S becomes =5101520
After the 2nd level, S=2550525100
Thus, we are able to make Target string So print YES
Input: S = “53783”, Target = “2893653”, X = 2, M = 3
Output: NO

方法：这是一个基于实现的问题，简单地从左到右遍历并在原始字符串中存在的每个数字中乘以X。使用给定的字符串S 执行此 M 次，并检查是否可以将原始字符串S转换为 Target 字符串。

下面是上述方法的实现：

C++

// C++ program to check
// If strings are Equal or not
 
#include 
using namespace std;
 
void makestring(string S, string Target, int X, int M)
{
    // Out of X and M if any one
    // having value zero then just compare
    // original string To Target string
    if (X == 0 || M == 0) {
        if (S == Target) {
            cout << "YES" << endl;
        }
    }
    else {
        vector v1;
        for (int i = 0; i < S.length(); i++) {
 
            // character to integer
            int val = S[i] - '0';
 
            // convert into string
            string a = to_string(val);
            v1.push_back(a);
        }
        while (M--) {
            vector v;
            for (int i = 0; i < v1.size(); i++) {
                for (int j = 0; j < v1[i].size(); j++) {
 
                    int temp = v1[i][j] - '0';
 
                    // multiply X into given integer
                    temp = temp * X;
 
                    v.push_back(to_string(temp));
                }
            }
            v1 = v;
        }
 
        // Store all character from vector
        // of string into temp string
        string temp = "";
        for (int i = 0; i < v1.size(); i++) {
            for (int j = 0; j < v1[i].size(); j++) {
                temp.push_back(v1[i][j]);
            }
        }
 
        // Compare both Target and temp
        if (temp == Target) {
            cout << "YES" << endl;
        }
        else {
            cout << "NO" << endl;
        }
    }
}
 
// Driver Code
int main()
{
    string S = "1234";
    string Target = "2550525100";
    int X = 5;
    int M = 2;
    makestring(S, Target, X, M);
    return 0;
}

Javascript

    // JavaScript program to check
    // If strings are Equal or not
    const makestring = (S, Target, X, M) => {
     
        // Out of X and M if any one
        // having value zero then just compare
        // original string To Target string
        if (X == 0 || M == 0) {
            if (S == Target) {
                document.write("YES<br/>");
            }
        }
        else {
            let v1 = [];
            for (let i = 0; i < S.length; i++) {
 
                // character to integer
                let val = S.charCodeAt(i) - '0'.charCodeAt(0);
 
                // convert into string
                let a = (val).toString();
                v1.push(a);
            }
            while (M--) {
                let v = [];
                for (let i = 0; i < v1.length; i++) {
                    for (let j = 0; j < v1[i].length; j++) {
 
                        let temp = v1[i].charCodeAt(j) - '0'.charCodeAt(0);
 
                        // multiply X into given integer
                        temp = temp * X;
 
                        v.push(temp.toString());
                    }
                }
                v1 = [...v];
            }
 
            // Store all character from vector
            // of string into temp string
            let temp = "";
            for (let i = 0; i < v1.length; i++) {
                for (let j = 0; j < v1[i].length; j++) {
                    temp += v1[i][j];
                }
            }
 
            // Compare both Target and temp
            if (temp == Target) {
                document.write("YES<br/>");
            }
            else {
                document.write("NO<br/>");
            }
        }
    }
 
    // Driver Code
    let S = "1234";
    let Target = "2550525100";
    let X = 5;
    let M = 2;
 
    makestring(S, Target, X, M);
 
    // This code is contributed by rakeshsahni
 

输出

YES

时间复杂度： O(N* N* M)
辅助空间： O(N)