给定一个正整数N，接着在X的升序一个矩阵V [] []，包括对{X，Y}，任务是提供一种用于在X的升序给出每一对中，可以执行以下操作：
- 将对{X，Y}转换为{X – Y，X}
- 将对{X，Y}转换为{X，X + Y}
- 将该对更改为{X，X}
任务是找到所需的加法和减法运算的计数，以确保没有两对重叠。
例子：
Input: N = 5, V[] = {{1, 2} {2, 1} {5, 10} {10, 9} {19, 1}}
Output: 3
Explanation:
{1, 2}: Operation 1 modifies pair to {-1, 1}.
{2, 1}: Operation 2 modifies pair to {2, 3}.
{5, 10}: Operation 3 modifies pair to {5, 5}
{10, 9}: Operation 3 modifies pair to {10, 10}
{19, 1}: Operation 2 modifies pair to {19, 20}.
Therefore, none of the pairs overlap. Hence, the count of addition and subtraction operations required is 3.
Input: N = 3, V[][] = {{10, 20} {15, 10} {20, 16}}
Output: 2
方法：
主要思想是观察到，无论如何，答案都不会超过N ，因为这三个操作中的任何一个都不能在一个对上进行两次。请按照以下步骤解决问题：
- 第一对始终选择“操作1 ”，因为第一对X最小。
- 始终选择最后一个对的操作2 ，因为X是最后一个对的最大值。
- 对于其余的对，检查应用操作1是否违反规则。如果不违反规则，那么它将始终使结果最大化。否则，请检查操作2 。如果两个操作中的任何一个适用，则增加计数。
- 如果这两个规则都不适用，请执行操作3。
- 最后，打印计数。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to find maximum count of operations
int find_max(vector > v, int n)
{
// Initialize count by 0
int count = 0;
if (n >= 2)
count = 2;
else
count = 1;
// Iterate over remaining pairs
for (int i = 1; i < n - 1; i++) {
// Check if first operation
// is applicable
if (v[i - 1].first
< (v[i].first - v[i].second))
count++;
// Check if 2nd operation is applicable
else if (v[i + 1].first
> (v[i].first + v[i].second)) {
count++;
v[i].first = v[i].first + v[i].second;
}
// Otherwise
else
continue;
}
// Return the count of operations
return count;
}
// Driver Code
int main()
{
int n = 3;
vector > v;
v.push_back({ 10, 20 });
v.push_back({ 15, 10 });
v.push_back({ 20, 16 });
cout << find_max(v, n);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find maximum count of operations
static int find_max(Vector v, int n)
{
// Initialize count by 0
int count = 0;
if (n >= 2)
count = 2;
else
count = 1;
// Iterate over remaining pairs
for(int i = 1; i < n - 1; i++)
{
// Check if first operation
// is applicable
if (v.get(i - 1).first <
(v.get(i).first - v.get(i).second))
count++;
// Check if 2nd operation is applicable
else if (v.get(i + 1).first >
(v.get(i).first + v.get(i).second))
{
count++;
v.get(i).first = v.get(i).first +
v.get(i).second;
}
// Otherwise
else
continue;
}
// Return the count of operations
return count;
}
// Driver Code
public static void main(String[] args)
{
int n = 3;
Vector v = new Vector<>();
v.add(new pair(10, 20));
v.add(new pair(15, 10));
v.add(new pair(20, 16));
System.out.print(find_max(v, n));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to implement
# the above approach
# Function to find maximum count of
# operations
def find_max(v, n):
# Initialize count by 0
count = 0
if(n >= 2):
count = 2
else:
count = 1
# Iterate over remaining pairs
for i in range(1, n - 1):
# Check if first operation
# is applicable
if(v[i - 1][0] > (v[i][0] +
v[i][1])):
count += 1
# Check if 2nd operation is applicable
elif(v[i + 1][0] > (v[i][0] +
v[i][1])):
count += 1
v[i][0] = v[i][0] + v[i][1]
# Otherwise
else:
continue
# Return the count of operations
return count
# Driver Code
n = 3
v = []
v.append([ 10, 20 ])
v.append([ 15, 10 ])
v.append([ 20, 16 ])
print(find_max(v, n))
# This code is contributed by Shivam SinghC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find maximum count of operations
static int find_max(List v, int n)
{
// Initialize count by 0
int count = 0;
if (n >= 2)
count = 2;
else
count = 1;
// Iterate over remaining pairs
for(int i = 1; i < n - 1; i++)
{
// Check if first operation
// is applicable
if (v[i - 1].first <
(v[i].first - v[i].second))
count++;
// Check if 2nd operation is applicable
else if (v[i + 1].first >
(v[i].first + v[i].second))
{
count++;
v[i].first = v[i].first +
v[i].second;
}
// Otherwise
else
continue;
}
// Return the count of operations
return count;
}
// Driver Code
public static void Main(String[] args)
{
int n = 3;
List v = new List();
v.Add(new pair(10, 20));
v.Add(new pair(15, 10));
v.Add(new pair(20, 16));
Console.Write(find_max(v, n));
}
}
// This code is contributed by 29AjayKumar 输出：
2
时间复杂度： O(N)
辅助空间： O(1)