使用给定数字达到目标的最小求和和模运算次数

给定一个数字N 、一个数组arr[]和一个目标数字K ，任务是通过选择任何数组元素并将每个元素中的N更改为(N + arr[i]) mod 100000来找到达到K的最小移动次数移动。

例子：

Input: N = 99880, K = 89, arr = {100, 3}
Output: 5
Explanation: Number “100” is used 2 times and the number “3” is pushed 3 times, resulting in K from N

Input: N = 10000, K = 10004, arr = {1}
Output: 4

方法：上述问题可以用动态规划解决，因为它有重叠的子问题和最优子结构。初始化dp表，其中dp[i]存储在 N 上达到状态i所需的最小移动，而dp[target]将是所需的答案。从每个arr[i] 中，找到所有可达状态并最小化dp值。请按照以下步骤操作：

• 初始化一个数组，比如dp[]来存储达到状态i所需的最小移动。
• 迭代数组 arr 并检查当前数字 N 的每个位置为x ：
  • 如果dp[x]等于-1，则继续。
  • 否则，迭代 while 循环直到dp[next]等于-1或dp[next]大于dp[x] + 1并更新next为(x+buttons[i])%100000和dp[next]为dp[ x]+ 1 。
• 最后，返回dp[target]的值。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to find the minimum moves
// to reach K from N
int minPushes(int N, int K, vector arr)
{
 
    // Intialization of dp vector
    vector dp(100000, -1);
 
    // dp[i] = minimum pushes
    // required to reach i
    dp[N] = 0;
 
    // Traversing through the buttons
    for (int i = 0; i < arr.size(); i++) {
 
        // Iterating through all the positions
        for (int xx = 0; xx < 100000; xx++) {
            int x = xx;
 
            // If not visited
            if (dp[x] == -1)
                continue;
 
            // Next status of lock
            int next = (x + arr[i]) % 100000;
            while (dp[next] == -1
                   || dp[next] > dp[x] + 1) {
                dp[next] = dp[x] + 1;
 
                // Advance to next state
                x = next;
                next = (next + arr[i]) % 100000;
            }
        }
    }
 
    // Return the final dp[target]
    return dp[K];
}
 
// Driver function
int main()
{
    // Given Input
    int N = 99880, K = 89;
    vector arr{ 100, 3 };
 
    // Function Call
    cout << minPushes(N, K, arr);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
 
// Function to find the minimum moves
// to reach K from N
static int minPushes(int N, int K, int[] arr)
{
 
    // Intialization of dp vector
    int[] dp = new int[100000];
    for (int i = 0; i < dp.length; i++)
        dp[i] = -1;
   
    // dp[i] = minimum pushes
    // required to reach i
    dp[N] = 0;
 
    // Traversing through the buttons
    for (int i = 0; i < arr.length; i++) {
 
        // Iterating through all the positions
        for (int xx = 0; xx < 100000; xx++) {
            int x = xx;
 
            // If not visited
            if (dp[x] == -1)
                continue;
 
            // Next status of lock
            int next = (x + arr[i]) % 100000;
            while (dp[next] == -1
                   || dp[next] > dp[x] + 1) {
                dp[next] = dp[x] + 1;
 
                // Advance to next state
                x = next;
                next = (next + arr[i]) % 100000;
            }
        }
    }
 
    // Return the final dp[target]
    return dp[K];
}
 
// Driver function
public static void main(String[] args)
{
    // Given Input
    int N = 99880, K = 89;
    int[] arr = { 100, 3 };
 
    // Function Call
    System.out.print(minPushes(N, K, arr));
 
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
        // JavaScript program for the above approach;
 
        // Function to find the minimum moves
        // to reach K from N
        function minPushes(N, K, arr) {
 
            // Intialization of dp vector
            let dp = new Array(100000).fill(-1);
 
            // dp[i] = minimum pushes
            // required to reach i
            dp[N] = 0;
 
            // Traversing through the buttons
            for (let i = 0; i < arr.length; i++) {
 
                // Iterating through all the positions
                for (let xx = 0; xx < 100000; xx++) {
                    let x = xx;
 
                    // If not visited
                    if (dp[x] == -1)
                        continue;
 
                    // Next status of lock
                    let next = (x + arr[i]) % 100000;
                    while (dp[next] == -1
                        || dp[next] > dp[x] + 1) {
                        dp[next] = dp[x] + 1;
 
                        // Advance to next state
                        x = next;
                        next = (next + arr[i]) % 100000;
                    }
                }
            }
 
            // Return the final dp[target]
            return dp[K];
        }
 
        // Driver function
 
        // Given Input
        let N = 99880, K = 89;
        let arr = [100, 3];
 
        // Function Call
        document.write(minPushes(N, K, arr));
 
 
   // This code is contributed by Potta Lokesh
    

C#

// C# implementation of the above approach
using System;
 
public class GFG{
 
// Function to find the minimum moves
// to reach K from N
static int minPushes(int N, int K, int[] arr)
{
 
    // Intialization of dp vector
    int[] dp = new int[100000];
    for (int i = 0; i < dp.Length; i++)
        dp[i] = -1;
   
    // dp[i] = minimum pushes
    // required to reach i
    dp[N] = 0;
 
    // Traversing through the buttons
    for (int i = 0; i < arr.Length; i++) {
 
        // Iterating through all the positions
        for (int xx = 0; xx < 100000; xx++) {
            int x = xx;
 
            // If not visited
            if (dp[x] == -1)
                continue;
 
            // Next status of lock
            int next = (x + arr[i]) % 100000;
            while (dp[next] == -1
                   || dp[next] > dp[x] + 1) {
                dp[next] = dp[x] + 1;
 
                // Advance to next state
                x = next;
                next = (next + arr[i]) % 100000;
            }
        }
    }
 
    // Return the readonly dp[target]
    return dp[K];
}
 
// Driver function
public static void Main(String[] args)
{
    // Given Input
    int N = 99880, K = 89;
    int[] arr = { 100, 3 };
 
    // Function Call
    Console.Write(minPushes(N, K, arr));
 
}
}
 
// This code is contributed by 29AjayKumar

输出

5

时间复杂度： O(N*M)
辅助空间： O(N)