给定大小为N的数组arr [] 。每秒钟，整数消失N秒，并在N秒后重新出现在其原始位置。整数从左到右依次消失arr [0]，arr [1]，…，arr [N – 1] 。在所有整数消失之后，它们开始重新出现，直到所有整数重新出现。一旦再次显示N个元素，该过程将再次开始。
现在给定Q个查询，每个查询由两个整数t和M组成。的任务是确定的M个从左边第i个元素在t个第二。如果直到M都不存在该数组，则打印-1 。
例子：
Input: arr[] = {1, 2, 3, 4, 5}, Q = {{1, 4}, {6, 1}, {3, 5}}
Output:
5
1
-1
At time,
t1 -> {2, 3, 4, 5}
t2 -> {3, 4, 5}
t3 -> {4, 5}
t4 -> {5}
t5 -> {}
t6 -> {1}
Input: arr[] = {5, 4, 3, 4, 5}, Q = {{2, 3}, {100000000, 2}}
Output:
5
4
方法：主要方法是需要检查数组是否为空或已满，并且可以通过将匝数除以数组大小来查看。如果余数为0，则可以是以下任一情况(空或填充)。
通过观察，可以看出该数组在奇数转弯中减小，而在偶数转弯中，该数组扩展，并且使用该观察结果，将检查M是否在索引之外或在数组内部。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to perform the queries
void PerformQueries(vector& a,
vector >& vec)
{
vector ans;
// Size of the array with
// 1-based indexing
int n = (int)a.size() - 1;
// Number of queries
int q = (int)vec.size();
// Iterating through the queries
for (int i = 0; i < q; ++i) {
long long t = vec[i].first;
int m = vec[i].second;
// If m is more than the
// size of the array
if (m > n) {
ans.push_back(-1);
continue;
}
// Count of turns
int turn = t / n;
// Find the remainder
int rem = t % n;
// If the remainder is 0 and turn is
// odd then the array is empty
if (rem == 0 and turn % 2 == 1) {
ans.push_back(-1);
continue;
}
// If the remainder is 0 and turn is
// even then array is full and
// is in its initial state
if (rem == 0 and turn % 2 == 0) {
ans.push_back(a[m]);
continue;
}
// If the remainder is not 0
// and the turn is even
if (turn % 2 == 0) {
// Current size of the array
int cursize = n - rem;
if (cursize < m) {
ans.push_back(-1);
continue;
}
ans.push_back(a[m + rem]);
}
else {
// Current size of the array
int cursize = rem;
if (cursize < m) {
ans.push_back(-1);
continue;
}
ans.push_back(a[m]);
}
}
// Print the result
for (int i : ans)
cout << i << "\n";
}
// Driver code
int main()
{
// The intial array, -1 is for
// 1 base indexing
vector a = { -1, 1, 2, 3, 4, 5 };
// Queries in the form of the pairs of (t, M)
vector > vec = {
{ 1, 4 },
{ 6, 1 },
{ 3, 5 }
};
PerformQueries(a, vec);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to perform the queries
static void PerformQueries(int[] a, int[][] vec)
{
Vector ans = new Vector<>();
// Size of the array with
// 1-based indexing
int n = (int) a.length - 1;
// Number of queries
int q = (int) vec.length;
// Iterating through the queries
for (int i = 0; i < q; ++i)
{
long t = vec[i][0];
int m = vec[i][1];
// If m is more than the
// size of the array
if (m > n)
{
ans.add(-1);
continue;
}
// Count of turns
int turn = (int) (t / n);
// Find the remainder
int rem = (int) (t % n);
// If the remainder is 0 and turn is
// odd then the array is empty
if (rem == 0 && turn % 2 == 1)
{
ans.add(-1);
continue;
}
// If the remainder is 0 and turn is
// even then array is full and
// is in its initial state
if (rem == 0 && turn % 2 == 0)
{
ans.add(a[m]);
continue;
}
// If the remainder is not 0
// and the turn is even
if (turn % 2 == 0)
{
// Current size of the array
int cursize = n - rem;
if (cursize < m)
{
ans.add(-1);
continue;
}
ans.add(a[m + rem]);
}
else
{
// Current size of the array
int cursize = rem;
if (cursize < m)
{
ans.add(-1);
continue;
}
ans.add(a[m]);
}
}
// Print the result
for (int i : ans)
System.out.print(i + "\n");
}
// Driver code
public static void main(String[] args)
{
// The intial array, -1 is for
// 1 base indexing
int[] a = { -1, 1, 2, 3, 4, 5 };
// Queries in the form of the pairs of (t, M)
int[][] vec = { { 1, 4 }, { 6, 1 }, { 3, 5 } };
PerformQueries(a, vec);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# Function to perform the queries
def PerformQueries(a, vec) :
ans = [];
# Size of the array with
# 1-based indexing
n = len(a) - 1;
# Number of queries
q = len(vec);
# Iterating through the queries
for i in range(q) :
t = vec[i][0];
m = vec[i][1];
# If m is more than the
# size of the array
if (m > n) :
ans.append(-1);
continue;
# Count of turns
turn = t // n;
# Find the remainder
rem = t % n;
# If the remainder is 0 and turn is
# odd then the array is empty
if (rem == 0 and turn % 2 == 1) :
ans.append(-1);
continue;
# If the remainder is 0 and turn is
# even then array is full and
# is in its initial state
if (rem == 0 and turn % 2 == 0) :
ans.append(a[m]);
continue;
# If the remainder is not 0
# and the turn is even
if (turn % 2 == 0) :
# Current size of the array
cursize = n - rem;
if (cursize < m) :
ans.append(-1);
continue;
ans.append(a[m + rem]);
else :
# Current size of the array
cursize = rem;
if (cursize < m) :
ans.append(-1);
continue;
ans.append(a[m]);
# Print the result
for i in ans :
print(i) ;
# Driver code
if __name__ == "__main__" :
# The intial array, -1 is for
# 1 base indexing
a = [ -1, 1, 2, 3, 4, 5 ];
# Queries in the form of the pairs of (t, M)
vec = [
[ 1, 4 ],
[ 6, 1 ],
[ 3, 5 ]
];
PerformQueries(a, vec);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to perform the queries
static void PerformQueries(int[] a, int[,] vec)
{
List ans = new List();
// Size of the array with
// 1-based indexing
int n = (int) a.Length - 1;
// Number of queries
int q = (int) vec.GetLength(0);
// Iterating through the queries
for (int i = 0; i < q; ++i)
{
long t = vec[i, 0];
int m = vec[i, 1];
// If m is more than the
// size of the array
if (m > n)
{
ans.Add(-1);
continue;
}
// Count of turns
int turn = (int) (t / n);
// Find the remainder
int rem = (int) (t % n);
// If the remainder is 0 and turn is
// odd then the array is empty
if (rem == 0 && turn % 2 == 1)
{
ans.Add(-1);
continue;
}
// If the remainder is 0 and turn is
// even then array is full and
// is in its initial state
if (rem == 0 && turn % 2 == 0)
{
ans.Add(a[m]);
continue;
}
// If the remainder is not 0
// and the turn is even
if (turn % 2 == 0)
{
// Current size of the array
int cursize = n - rem;
if (cursize < m)
{
ans.Add(-1);
continue;
}
ans.Add(a[m + rem]);
}
else
{
// Current size of the array
int cursize = rem;
if (cursize < m)
{
ans.Add(-1);
continue;
}
ans.Add(a[m]);
}
}
// Print the result
foreach (int i in ans)
Console.Write(i + "\n");
}
// Driver code
public static void Main(String[] args)
{
// The intial array, -1 is for
// 1 base indexing
int[] a = { -1, 1, 2, 3, 4, 5 };
// Queries in the form of the pairs of (t, M)
int[,] vec = { { 1, 4 }, { 6, 1 }, { 3, 5 } };
PerformQueries(a, vec);
}
}
// This code is contributed by 29AjayKumar 输出：
5
1
-1