给定两个由正整数和负整数(包括0)组成的N个整数序列(序列A和序列B)。然后会有Q个查询，在每个查询中，您将得到两个整数l和r(r> = l)。让我们定义一个函数：
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任务是为每个查询打印Max(f(x(y，y)) – Min(f(x(y，y)))的值。
例子：
Input:
N = 5, Q = 2
A[] = 0 7 3 4 5
B[] = 0 3 1 2 3
l = 1, r = 1
l = 1, r = 3
Output:
0
917448
Input:
N = 5, Q = 2
A[] = 0 -8 3 4 -9
B[] = 0 -3 -5 2 3
l = 1, r = 1
l = 1, r = 3
Output:
0
851916
方法：如果我们清楚地看到函数仅取决于x和y的值，则可以通过查看各个数组的总和的性质轻松地计算出x和y的值。首先，我们预先计算两个数组的前缀和，然后我们就可以看到求和的性质。并且x和y的值始终为32766或-32766，因为我们采用它们的模。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
#define ll long long
#define MAX 200006
#define CONS 32766
// Function to calculate the value
void calc(ll a[], ll b[], ll lr[], ll q, ll n)
{
ll M, m, i, j, k, l, r, suma, sumb, cc;
cc = 0;
// forming the prefix sum arrays
for (i = 0; i < n - 1; ++i) {
a[i + 1] += a[i];
b[i + 1] += b[i];
}
while (q--) {
// Taking the query
l = lr[cc++];
r = lr[cc++];
l -= 2;
r -= 1;
// finding the sum in the range l to r in array a
suma = a[r];
// finding the sum in the range l to r in array b
sumb = b[r];
if (l >= 0) {
suma -= a[l];
sumb -= b[l];
}
// Finding the max value of the function
M = max(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
M = max(M, max(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
// Finding the min value of the function
m = min(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
m = min(m, min(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
cout << (M - m) << "\n";
}
}
// Driver code
int main()
{
ll n = 5, q = 2;
ll a[5] = { 0, 7, 3, 4, 5 };
ll b[5] = { 0, 3, 1, 2, 3 };
ll lr[q * 2];
lr[0] = 1;
lr[1] = 1;
lr[2] = 1;
lr[3] = 3;
calc(a, b, lr, q, n);
return 0;
} Java
// Java implementation of above approach
import java.util.*;
class GFG
{
static final int MAX=200006;
static final int CONS=32766;
// Function to calculate the value
static void calc(int a[], int b[], int lr[], int q, int n)
{
int M, m, i, j, k, l, r, suma, sumb, cc;
cc = 0;
// forming the prefix sum arrays
for (i = 0; i < n - 1; ++i)
{
a[i + 1] += a[i];
b[i + 1] += b[i];
}
while (q!=0)
{
// Taking the query
l = lr[cc++];
r = lr[cc++];
l -= 2;
r -= 1;
// finding the sum in the range l to r in array a
suma = a[r];
// finding the sum in the range l to r in array b
sumb = b[r];
if (l >= 0) {
suma -= a[l];
sumb -= b[l];
}
// Finding the max value of the function
M = Math.max(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
M = Math.max(M, Math.max(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
// Finding the min value of the function
m = Math.min(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
m = Math.min(m, Math.min(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
System.out.println((M - m) );
q--;
}
}
// Driver code
public static void main(String [] args)
{
int n = 5, q = 2;
int []a = { 0, 7, 3, 4, 5 };
int []b = { 0, 3, 1, 2, 3 };
int []lr=new int[q * 2];
lr[0] = 1;
lr[1] = 1;
lr[2] = 1;
lr[3] = 3;
calc(a, b, lr, q, n);
}
// This code is contributed by ihritik
}Python3
# Python 3 implementation of
# above approach
MAX = 200006
CONS = 32766
# Function to calculate the value
def calc(a, b, lr, q, n):
cc = 0
# forming the prefix sum arrays
for i in range(n - 1) :
a[i + 1] += a[i]
b[i + 1] += b[i]
while (q > 0) :
# Taking the query
l = lr[cc]
cc +=1
r = lr[cc]
cc += 1
l -= 2
r -= 1
# finding the sum in the range l
# to r in array a
suma = a[r]
# finding the sum in the range
# l to r in array b
sumb = b[r]
if (l >= 0) :
suma -= a[l]
sumb -= b[l]
# Finding the max value of the function
M = max(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb)
M = max(M, max(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb))
# Finding the min value of the function
m = min(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb)
m = min(m, min(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb))
print(M - m)
q -= 1
# Driver code
if __name__ == "__main__":
n = 5
q = 2
a = [ 0, 7, 3, 4, 5 ]
b = [ 0, 3, 1, 2, 3 ]
lr = [0]*(q * 2)
lr[0] = 1
lr[1] = 1
lr[2] = 1
lr[3] = 3
calc(a, b, lr, q, n)
# This code is contributed by
# ChitraNayalC#
// C# implementation of above approach
using System;
class GFG
{
// static int MAX=200006;
static int CONS = 32766;
// Function to calculate the value
static void calc(int []a, int []b,
int []lr, int q, int n)
{
int M, m, i, l, r, suma, sumb, cc;
cc = 0;
// forming the prefix sum arrays
for (i = 0; i < n - 1; ++i)
{
a[i + 1] += a[i];
b[i + 1] += b[i];
}
while (q != 0)
{
// Taking the query
l = lr[cc++];
r = lr[cc++];
l -= 2;
r -= 1;
// finding the sum in the
// range l to r in array a
suma = a[r];
// finding the sum in the
// range l to r in array b
sumb = b[r];
if (l >= 0)
{
suma -= a[l];
sumb -= b[l];
}
// Finding the max value of the function
M = Math.Max(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
M = Math.Max(M, Math.Max(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
// Finding the min value of the function
m = Math.Min(CONS * suma + CONS * sumb,
-CONS * suma - CONS * sumb);
m = Math.Min(m, Math.Min(CONS * suma - CONS * sumb,
-CONS * suma + CONS * sumb));
Console.WriteLine((M - m));
q--;
}
}
// Driver code
public static void Main()
{
int n = 5, q = 2;
int []a = { 0, 7, 3, 4, 5 };
int []b = { 0, 3, 1, 2, 3 };
int []lr = new int[q * 2];
lr[0] = 1;
lr[1] = 1;
lr[2] = 1;
lr[3] = 3;
calc(a, b, lr, q, n);
}
}
// This code is contributed by anuj_67PHP
= 0)
{
$suma -= $a[$l];
$sumb -= $b[$l];
}
// Finding the max value of the function
$M = max($CONS * $suma + $CONS * $sumb,
-$CONS * $suma - $CONS * $sumb);
$M = max($M, max($CONS * $suma - $CONS * $sumb,
-$CONS * $suma + $CONS * $sumb));
// Finding the min value of the function
$m = min($CONS * $suma + $CONS * $sumb,
-$CONS * $suma - $CONS * $sumb);
$m = min($m, min($CONS * $suma - $CONS * $sumb,
-$CONS * $suma + $CONS * $sumb));
echo ($M - $m) , "\n";
}
}
// Driver code
$n = 5; $q = 2;
$a = array(0, 7, 3, 4, 5 );
$b = array( 0, 3, 1, 2, 3 );
$lr[0] = 1;
$lr[1] = 1;
$lr[2] = 1;
$lr[3] = 3;
calc($a, $b, $lr, $q, $n);
// This code is contributed by anuj_67
?>输出：
0
917448
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