给定N个元素和两个初始值a和b的二进制数组。如果满足以下条件，我们可以越过第i个元素：
- 如果a [i] == 0 ，那么我们可以使用b或a中的1个单位穿过第i个元素。
- 如果a [i] == 1 ，则如果我们使用b中的1个单位，则a增加1个单位。如果从a使用1个单位，则a或b都不会增加。
任务是找到使用a和b单位可以交叉的最大元素数。
注意：如果我们将a任意增加1，则它不能超过a的原始值。
例子：
Input: arr[] = {0, 1, 0, 1, 0}, a = 1, b = 2;
Output: 5
Use 1 unit from a to cross 1st element. (a = 0 and b = 2)
Use 1 unit from b to cross 2nd element. (a = 1 and b = 1)
Use 1 unit from a to cross 3rd element. (a = 0 and b = 1)
Use 1 unit from b to cross 4th element. (a = 1 and b = 0)
Use 1 unit from a to cross 5th element. (a = 0 and b = 0)
Input: a[] = {1, 0, 0, 1, 0, 1}, a = 1, b = 2
Use 1 unit from b to cross first element. (a = 1 and b = 1)
Use 1 unit from b to cross second element. (a = 1 and b = 0)
Use 1 unit from a to cross third element. (a = 0 and b = 0)
Output: 3
方法：迭代数组元素并执行以下步骤：
- 如果没有b或a传递元素，则中断。
- 否则，如果没有剩余，则使用b；如果arr [i] == 1，则将a加1。
- 否则，如果没有b，则使用a。
- 否则，如果arr [i] == 1，则使用b并将a增加1直到原始a的最大值。
- 否则，只需使用a中的1个单位。
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 the number
// of elements crossed
int findElementsCrossed(int arr[], int a, int b, int n)
{
// Keep a copy of a
int aa = a;
int ans = 0;
// Iterate in the binary array
for (int i = 0; i < n; i++) {
// If no a and b left to use
if (a == 0 && b == 0)
break;
// If there is no a
else if (a == 0) {
// use b and increase a by 1
// if arr[i] is 1
if (arr[i] == 1) {
b -= 1;
a = min(aa, a + 1);
}
// simply use b
else
b -= 1;
}
// Use a if theres no b
else if (b == 0)
a--;
// Increase a and use b if arr[i] == 1
else if (arr[i] == 1 && a < aa) {
b -= 1;
a = min(aa, a + 1);
}
// Use a
else
a--;
ans++;
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 0, 0, 1, 0, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
int a = 1;
int b = 2;
cout << findElementsCrossed(arr, a, b, n);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find the number
// of elements crossed
static int findElementsCrossed(int arr[],
int a, int b, int n)
{
// Keep a copy of a
int aa = a;
int ans = 0;
// Iterate in the binary array
for (int i = 0; i < n; i++)
{
// If no a and b left to use
if (a == 0 && b == 0)
break;
// If there is no a
else if (a == 0)
{
// use b and increase a by 1
// if arr[i] is 1
if (arr[i] == 1)
{
b -= 1;
a = Math.min(aa, a + 1);
}
// simply use b
else
b -= 1;
}
// Use a if theres no b
else if (b == 0)
a--;
// Increase a and use b if arr[i] == 1
else if (arr[i] == 1 && a < aa)
{
b -= 1;
a = Math.min(aa, a + 1);
}
// Use a
else
a--;
ans++;
}
return ans;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 0, 0, 1, 0, 1 };
int n = arr.length;
int a = 1;
int b = 2;
System.out.println(findElementsCrossed(arr, a, b, n));
}
}
// This code is contributed by
// Surendra_GangwarPython3
# Python3 program to implement
# the above approach
# Function to find the number
# of elements crossed
def findElementsCrossed(arr, a, b, n):
# Keep a copy of a
aa = a
ans = 0
# Iterate in the binary array
for i in range(n):
# If no a and b left to use
if (a == 0 and b == 0):
break
# If there is no a
elif (a == 0):
# use b and increase a by 1
# if arr[i] is 1
if (arr[i] == 1):
b -= 1
a = min(aa, a + 1)
# simply use b
else:
b -= 1
# Use a if theres no b
elif (b == 0):
a -= 1
# Increase a and use b if arr[i] == 1
elif (arr[i] == 1 and a < aa):
b -= 1
a = min(aa, a + 1)
# Use a
else:
a -= 1
ans += 1
return ans
# Driver code
arr = [1, 0, 0, 1, 0, 1]
n = len(arr)
a = 1
b = 2
print(findElementsCrossed(arr, a, b, n))
# This code is contributed by mohit kumarC#
// C# implementation of the above approach
using System;
class GFG
{
// Function to find the number
// of elements crossed
static int findElementsCrossed(int []arr,
int a, int b, int n)
{
// Keep a copy of a
int aa = a;
int ans = 0;
// Iterate in the binary array
for (int i = 0; i < n; i++)
{
// If no a and b left to use
if (a == 0 && b == 0)
break;
// If there is no a
else if (a == 0)
{
// use b and increase a by 1
// if arr[i] is 1
if (arr[i] == 1)
{
b -= 1;
a = Math.Min(aa, a + 1);
}
// simply use b
else
b -= 1;
}
// Use a if theres no b
else if (b == 0)
a--;
// Increase a and use b if arr[i] == 1
else if (arr[i] == 1 && a < aa)
{
b -= 1;
a = Math.Min(aa, a + 1);
}
// Use a
else
a--;
ans++;
}
return ans;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 0, 0, 1, 0, 1 };
int n = arr.Length;
int a = 1;
int b = 2;
Console.WriteLine(findElementsCrossed(arr, a, b, n));
}
}
// This code contributed by Rajput-JiPHP
输出：
3