给定一个由N个整数和一个整数M组成的数组。您可以更改数组中任何元素的符号(正号或负号)。任务是打印数组元素的所有可能组合，这些组合可以通过更改元素的符号来获得，以使它们的和可被M整除。
注意：您必须采用每种组合中的所有数组元素，并以与数组中存在的元素相同的顺序。但是，您可以更改元素的符号。
例子：
Input: a[] = {5, 6, 7}, M = 3
Output:
-5-6-7
+5-6+7
-5+6-7
+5+6+7
Input: a[] = {3, 5, 6, 8}, M = 5
Output:
-3-5+6-8
-3+5+6-8
+3-5-6+8
+3+5-6+8
方法：此处使用功率设置的概念来解决此问题。使用功率集生成可以应用于元素数组的所有可能的符号组合。如果获得的总和可被M整除，则打印该组合。步骤如下：
- 使用幂集迭代“ +”和“-”的所有可能组合。
- 迭代数组元素，如果从左数第j位被设置，则假定该数组元素为正；如果未设置该位，则假定该数组元素为负。请参考此处检查是否设置了任何索引。
- 如果总和可被M整除，则再次遍历数组元素，并将它们与sign(’+’或’-‘)一起打印。
下面是上述方法的实现：
C++
#include
using namespace std;
// Function to print all the combinations
void printCombinations(int a[], int n, int m)
{
// Iterate for all combinations
for (int i = 0; i < (1 << n); i++) {
int sum = 0;
// Initially 100 in binary if n is 3
// as 1<<(3-1) = 100 in binary
int num = 1 << (n - 1);
// Iterate in the array and assign signs
// to the array elements
for (int j = 0; j < n; j++) {
// If the j-th bit from left is set
// take '+' sign
if (i & num)
sum += a[j];
else
sum += (-1 * a[j]);
// Right shift to check if
// jth bit is set or not
num = num >> 1;
}
if (sum % m == 0) {
// re-initialize
num = 1 << (n - 1);
// Iterate in the array elements
for (int j = 0; j < n; j++) {
// If the jth from left is set
if ((i & num))
cout << "+" << a[j] << " ";
else
cout << "-" << a[j] << " ";
// right shift
num = num >> 1;
}
cout << endl;
}
}
}
// Driver Code
int main()
{
int a[] = { 3, 5, 6, 8 };
int n = sizeof(a) / sizeof(a[0]);
int m = 5;
printCombinations(a, n, m);
return 0;
} Java
import java.io.*;
class GFG
{
// Function to print
// all the combinations
static void printCombinations(int a[],
int n, int m)
{
// Iterate for all
// combinations
for (int i = 0;
i < (1 << n); i++)
{
int sum = 0;
// Initially 100 in binary
// if n is 3 as
// 1<<(3-1) = 100 in binary
int num = 1 << (n - 1);
// Iterate in the array
// and assign signs to
// the array elements
for (int j = 0; j < n; j++)
{
// If the j-th bit
// from left is set
// take '+' sign
if ((i & num) > 0)
sum += a[j];
else
sum += (-1 * a[j]);
// Right shift to check if
// jth bit is set or not
num = num >> 1;
}
if (sum % m == 0)
{
// re-initialize
num = 1 << (n - 1);
// Iterate in the
// array elements
for (int j = 0; j < n; j++)
{
// If the jth from
// left is set
if ((i & num) > 0)
System.out.print("+" +
a[j] + " ");
else
System.out.print("-" +
a[j] + " ");
// right shift
num = num >> 1;
}
System.out.println();
}
}
}
// Driver code
public static void main(String args[])
{
int a[] = { 3, 5, 6, 8 };
int n = a.length;
int m = 5;
printCombinations(a, n, m);
}
}
// This code is contributed
// by inder_verma.Python3
# Function to print
# all the combinations
def printCombinations(a, n, m):
# Iterate for all
# combinations
for i in range(0, (1 << n)):
sum = 0
# Initially 100 in binary
# if n is 3 as
# 1<<(3-1) = 100 in binary
num = 1 << (n - 1)
# Iterate in the array
# and assign signs to
# the array elements
for j in range(0, n):
# If the j-th bit
# from left is set
# take '+' sign
if ((i & num) > 0):
sum += a[j]
else:
sum += (-1 * a[j])
# Right shift to check if
# jth bit is set or not
num = num >> 1
if (sum % m == 0):
# re-initialize
num = 1 << (n - 1)
# Iterate in the
# array elements
for j in range(0, n):
# If the jth from
# left is set
if ((i & num) > 0):
print("+", a[j], end = " ",
sep = "")
else:
print("-", a[j], end = " ",
sep = "")
# right shift
num = num >> 1
print("")
# Driver code
a = [ 3, 5, 6, 8 ]
n = len(a)
m = 5
printCombinations(a, n, m)
# This code is contributed
# by smita.C#
// Print all the combinations
// of N elements by changing
// sign such that their sum
// is divisible by M
using System;
class GFG
{
// Function to print
// all the combinations
static void printCombinations(int []a,
int n, int m)
{
// Iterate for all
// combinations
for (int i = 0;
i < (1 << n); i++)
{
int sum = 0;
// Initially 100 in binary
// if n is 3 as
// 1<<(3-1) = 100 in binary
int num = 1 << (n - 1);
// Iterate in the array
// and assign signs to
// the array elements
for (int j = 0; j < n; j++)
{
// If the j-th bit
// from left is set
// take '+' sign
if ((i & num) > 0)
sum += a[j];
else
sum += (-1 * a[j]);
// Right shift to check if
// jth bit is set or not
num = num >> 1;
}
if (sum % m == 0)
{
// re-initialize
num = 1 << (n - 1);
// Iterate in the
// array elements
for (int j = 0; j < n; j++)
{
// If the jth from
// left is set
if ((i & num) > 0)
Console.Write("+" +
a[j] + " ");
else
Console.Write("-" +
a[j] + " ");
// right shift
num = num >> 1;
}
Console.Write("\n");
}
}
}
// Driver code
public static void Main()
{
int []a = { 3, 5, 6, 8 };
int n = a.Length;
int m = 5;
printCombinations(a, n, m);
}
}
// This code is contributed
// by Smitha.PHP
> 1;
}
if ($sum % $m == 0)
{
// re-initialize
$num = 1 << ($n - 1);
// Iterate in the array elements
for ($j = 0; $j < $n; $j++)
{
// If the jth from left is set
if (($i & $num))
echo "+" , $a[$j] , " ";
else
echo "-" , $a[$j] , " ";
// right shift
$num = $num >> 1;
}
echo "\n";
}
}
}
// Driver Code
$a = array( 3, 5, 6, 8 );
$n = sizeof($a);
$m = 5;
printCombinations($a, $n, $m);
// This code is contributed by ajit
?>输出：
-3 -5 +6 -8
-3 +5 +6 -8
+3 -5 -6 +8
+3 +5 -6 +8
时间复杂度： O(2 N * N)，其中N是元素数。