在 Q 查询更新后打印大小为 N 的数组，其中包含 [0, M) 范围内的值
给定 大小为N的数组arr[]包含状态从0到 ( M – 1)的循环变量（即，当从M-1递增时，它变为0 ）。任务是完成以下两种类型中的任何一种的Q查询：
- 第一种类型：1 LRK – 循环增加索引范围内的所有值 [L, R]，K 次。
- 第二种类型：2 LR - 打印索引范围内的更新值 [L, R]。
例子：
Input: arr[] = {2, 2, 7, 2, 5}, Q = 5, M = 8
queries[][] = {{1, 0, 3, 4},
{1, 4, 4, 2},
{1, 0, 0, 7},
{2, 1, 3},
{2, 3, 3}}
Output: {6, 3, 6}, {6}
Explanation: The states after performing each operation are:
After 1st: {6, 6, 3, 6, 5}
After 2nd: {6, 6, 3, 6, 7}
After 3rd: {5, 6, 3, 6, 7}
So in 4th query elements from index 1 to 3 and in 5th query element in index 3 are printed.
Input: arr[] = [2, 3, 4, 5], Q = 3, M = 6
queries[][] = {{1, 0, 0, 3},
{1, 1, 2, 2},
{1, 3, 3, 1},
{2, 0, 3}}
Output: {5, 5, 1, 1}
方法：可以使用贪心方法解决问题。按照下面提到的步骤来实施该方法：
- 当查询是第一种类型时，更新范围 [L, R] 中的所有值，K 次。
- 当查询是第二种类型时，打印范围 [L, R] 中的值。
下面是上述方法的实现。
C++14
// C++ code to implement above approach
#include
using namespace std;
// Function to implement the queries
void update(int arr[], int N, int M, int Q,
vector >& queries)
{
// Loop to implement the queries
for (int i = 0; i < Q; i++) {
if (queries[i][0] == 1) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
arr[j] = (arr[j] +
queries[i][3]) % M;
}
else if (queries[i][0] == 2) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
cout << arr[j] << " ";
cout << endl;
}
}
}
// Driver's code
int main()
{
int N = 5, M = 8, Q = 5;
int arr[] = { 2, 2, 7, 2, 5 };
vector > queries(Q);
queries[0] = { 1, 0, 3, 4 };
queries[1] = { 1, 4, 4, 2 };
queries[2] = { 1, 0, 0, 7 };
queries[3] = { 2, 1, 3 };
queries[4] = { 2, 3, 3 };
update(arr, N, M, Q, queries);
return 0;
} Java
// Java code to implement the above approach
import java.io.*;
class GFG {
// Function to implement the queries
public static void update(int[] arr, int N, int M,
int Q, int[][] queries)
{
// Loop to implement the queries
for (int i = 0; i < Q; i++) {
if (queries[i][0] == 1) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
arr[j] = (arr[j] +
queries[i][3]) % M;
}
else if (queries[i][0] == 2) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
System.out.print(arr[j]+ " ");
System.out.println();
}
}
}
// Driver's code
public static void main (String[] args)
{
int N = 5, M = 8, Q = 5;
int[] arr = { 2, 2, 7, 2, 5 };
int[][] queries = new int[][]
{
new int[] { 1, 0, 3, 4 },
new int[] { 1, 4, 4, 2 },
new int[] { 1, 0, 0, 7 },
new int[] { 2, 1, 3 },
new int[] { 2, 3, 3 }
};
update(arr, N, M, Q, queries);
}
}
// This code is contributed by Shubham SinghPython3
# Python3 code to implement above approach
# Function to implement the queries
def update(arr, N, M, Q, queries):
# Loop to implement the queries
for i in range(Q):
if (queries[i][0] == 1):
for j in range(queries[i][1],
queries[i][2] + 1):
arr[j] = (arr[j] +
queries[i][3]) % M
elif (queries[i][0] == 2):
for j in range(queries[i][1],
queries[i][2] + 1):
print(arr[j], end = " ")
print()
# Driver code
if __name__ == "__main__":
N = 5
M = 8
Q = 5
arr = [2, 2, 7, 2, 5]
queries = []
queries.append([1, 0, 3, 4])
queries.append([1, 4, 4, 2])
queries.append([1, 0, 0, 7])
queries.append([2, 1, 3])
queries.append([2, 3, 3])
update(arr, N, M, Q, queries)
# This code is contributed by ukaspC#
// C# code to implement the above approach
using System;
public class GFG{
// Function to implement the queries
public static void update(int[] arr, int N, int M, int Q, int[][] queries)
{
// Loop to implement the queries
for (int i = 0; i < Q; i++) {
if (queries[i][0] == 1) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
arr[j] = (arr[j] +
queries[i][3]) % M;
}
else if (queries[i][0] == 2) {
for (int j = queries[i][1];
j <= queries[i][2];
j++)
Console.Write(arr[j]+ " ");
Console.WriteLine();
}
}
}
// Driver's code
public static void Main()
{
int N = 5, M = 8, Q = 5;
int[] arr = { 2, 2, 7, 2, 5 };
int[][] queries = new int[][]
{
new int[] { 1, 0, 3, 4 },
new int[] { 1, 4, 4, 2 },
new int[] { 1, 0, 0, 7 },
new int[] { 2, 1, 3 },
new int[] { 2, 3, 3 }
};
update(arr, N, M, Q, queries);
}
}
// This code is contributed by Shubham SinghJavascript
输出
6 3 6
6 时间复杂度： O(Q*N)
辅助空间： O(1)