有n台服务器。每个服务器i当前正在处理一个(i)数量的请求。还有另一个数组b，其中b(i)代表调度到服务器i的传入请求的数量。重新安排传入的请求，以使每个服务器i在重新安排之后都拥有相等数量的请求。到服务器i的传入请求只能重新安排到服务器i-1，i，i + 1。如果没有这种重新安排的可能性，则在重新安排之后，输出-1，否则每个服务器保留的请求的打印数量。
例子：
Input : a = {6, 14, 21, 1}
b = {15, 7, 10, 10}
Output : 21
b(0) scheduled to a(0) --> a(0) = 21
b(1) scheduled to a(1) --> a(1) = 21
b(2) scheduled to a(3) --> a(3) = 11
b(3) scheduled to a(3) --> a(3) = 21
a(2) remains unchanged --> a(2) = 21
Input : a = {1, 2, 3}
b = {1, 100, 3}
Output : -1
No rescheduling will result in equal requests.推荐：请首先在IDE上尝试您的方法，然后查看解决方案
方法：观察到数组b的每个元素总是总是恰好一次添加到数组a的任何一个元素。因此，数组b的所有元素的总和+旧数组a的所有元素的总和=新数组a的所有元素的总和。令该总和为S。新数组a的所有元素也相等。让每个新元素为x。如果数组a具有n个元素，则给出
x * n = S
=> x = S/n ....(1)因此，新数组a的所有相等元素都由eqn(1)给出。现在，要使每个a(i)等于x，我们需要将xa(i)添加到每个元素。我们将遍历整个数组a，并检查a(i)是否可以等于x。有多种可能性：
1. a(i)> x：在这种情况下，永远不能使a(i)等于x。因此输出-1。
2. a(i)+ b(i)+ b(i + 1)= x。只需将b(i)+ b(i + 1)添加到a(i)，然后将b(i)，b(i + 1)更新为零即可。
3. a(i)+ b(i)= x。将b(i)添加到a(i)，并将b(i)更新为零。
4. a(i)+ b(i + 1)= x。将b(i + 1)添加到a(i)，并将b(i + 1)更新为零。
数组a完全遍历后，检查数组b的所有元素是否为零。如果是，则打印a(0)，否则打印-1。
为什么b(i)加法后更新为零?
考虑一个测试案例，其中b(i)既不添加到a(i-1)也不添加到a(i)。在那种情况下，我们必须将b(i)添加到a(i + 1)。因此，当我们开始对元素a(i)执行计算时，在数组a上进行迭代时，首先我们将元素b(i-1)添加到a(i)来考虑上述可能性。现在，如果b(i-1)已被添加到a(i-1)或a(i-2)，那么在这种情况下，就无法将其添加到a(i)。因此，为避免b(i)的这种双重加法，将其更新为零。
逐步算法为：
1. Compute sum S and find x = S / n
2. Iterate over array a
3. for each element a(i) do:
a(i) += b(i-1)
b(i-1) = 0;
if a(i) > x:
break
else:
check for other three possibilities
and update a(i) and b(i).
4. Check whether all elements of b(i) are
zero or not.执行：
C++
// CPP program to schedule jobs so that
// each server gets equal load.
#include
using namespace std;
// Function to find new array a
int solve(int a[], int b[], int n)
{
int i;
long long int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++) {
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0) {
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x) {
a[i] = y;
b[i] = b[i + 1] = 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x) {
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x) {
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}
int main()
{
int a[] = { 6, 14, 21, 1 };
int b[] = { 15, 7, 10, 10 };
int n = sizeof(a) / sizeof(a[0]);
cout << solve(a, b, n);
return 0;
} Java
// Java program to schedule jobs so that
// each server gets equal load.
class GFG
{
// Function to find new array a
static int solve(int a[], int b[], int n)
{
int i;
int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++)
{
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0)
{
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x)
{
a[i] = y;
b[i]= 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x)
{
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x)
{
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 6, 14, 21, 1 };
int b[] = { 15, 7, 10, 10 };
int n = a.length;
System.out.println(solve(a, b, n));
}
}
// This code contributed by Rajput-JiPython3
# Python3 program to schedule jobs so that
# each server gets an equal load.
# Function to find new array a
def solve(a, b, n):
s = 0
# find sum S of both arrays a and b.
for i in range(0, n):
s += a[i] + b[i]
# Single element case.
if n == 1:
return a[0] + b[0]
# This checks whether sum s can be divided
# equally between all array elements. i.e.
# whether all elements can take equal value
# or not.
if s % n != 0:
return -1
# Compute possible value of new
# array elements.
x = s // n
for i in range(0, n):
# Possibility 1
if a[i] > x:
return -1
# ensuring that all elements of
# array b are used.
if i > 0:
a[i] += b[i - 1]
b[i - 1] = 0
# If a(i) already updated to x
# move to next element in array a.
if a[i] == x:
continue
# Possibility 2
y = a[i] + b[i]
if i + 1 < n:
y += b[i + 1]
if y == x:
a[i] = y
b[i] = 0
if i + 1 < n: b[i + 1] = 0
continue
# Possibility 3
if a[i] + b[i] == x:
a[i] += b[i]
b[i] = 0
continue
# Possibility 4
if i + 1 < n and a[i] + b[i + 1] == x:
a[i] += b[i + 1]
b[i + 1] = 0
continue
# If a(i) can not be made equal
# to x even after adding all
# possible elements from b(i)
# then print -1.
return -1
# check whether all elements of b
# are used.
for i in range(0, n):
if b[i] != 0:
return -1
# Return the new array element value.
return x
# Driver Code
if __name__ == "__main__":
a = [6, 14, 21, 1]
b = [15, 7, 10, 10]
n = len(a)
print(solve(a, b, n))
# This code is contributed by Rituraj JainC#
// C# program to schedule jobs so that
// each server gets equal load.
using System;
class GFG
{
// Function to find new array a
static int solve(int []a, int []b, int n)
{
int i;
int s = 0;
// find sum S of both arrays a and b.
for (i = 0; i < n; i++)
s += (a[i] + b[i]);
// Single element case.
if (n == 1)
return a[0] + b[0];
// This checks whether sum s can be divided
// equally between all array elements. i.e.
// whether all elements can take equal value
// or not.
if (s % n != 0)
return -1;
// Compute possible value of new array
// elements.
int x = s / n;
for (i = 0; i < n; i++)
{
// Possibility 1
if (a[i] > x)
return -1;
// ensuring that all elements of
// array b are used.
if (i > 0)
{
a[i] += b[i - 1];
b[i - 1] = 0;
}
// If a(i) already updated to x
// move to next element in array a.
if (a[i] == x)
continue;
// Possibility 2
int y = a[i] + b[i];
if (i + 1 < n)
y += b[i + 1];
if (y == x)
{
a[i] = y;
b[i]= 0;
continue;
}
// Possibility 3
if (a[i] + b[i] == x)
{
a[i] += b[i];
b[i] = 0;
continue;
}
// Possibility 4
if (i + 1 < n &&
a[i] + b[i + 1] == x)
{
a[i] += b[i + 1];
b[i + 1] = 0;
continue;
}
// If a(i) can not be made equal
// to x even after adding all
// possible elements from b(i)
// then print -1.
return -1;
}
// check whether all elements of b
// are used.
for (i = 0; i < n; i++)
if (b[i] != 0)
return -1;
// Return the new array element value.
return x;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 6, 14, 21, 1 };
int []b = { 15, 7, 10, 10 };
int n = a.Length;
Console.WriteLine(solve(a, b, n));
}
}
// This code has been contributed by 29AjayKumarJavascript
输出：
21时间复杂度： O(n)
辅助空间： O(1)如果不允许我们修改原始数组，则O(n)