给定N个数字，求出奇数索引处的元素之和与偶数索引处的元素之和相等的排列数目。
例子：
Input: 1 2 3
Output: 2
The permutations are:
1 3 2 sum at odd index = 1+2 = 3, sum at even index = 3
2 3 1 sum at odd index = 2+1 = 3, sum at even index = 3
Input: 1 2 1 2
Output: 3
The permutations are:
1 2 2 1
2 1 1 2
2 2 1 1
解决该问题的方法是在C++ STL中使用next_permutation()，这有助于生成N个数字的所有可能排列。如果奇数索引元素的总和等于生成的排列的偶数索引元素的总和，则增加计数。检查所有排列后，打印计数。
下面是上述方法的实现：
C++
// C++ program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
#include
using namespace std;
// Function that returns the number of permutations
int numberOfPermutations(int a[], int n)
{
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutatio
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2)
sumOdd += a[i];
else
sumEven += a[i];
}
// If condition holds
if (sumOdd == sumEven)
c++;
} while (next_permutation(a, a + n));
// return the number of permutations
return c;
}
// Driver Code
int main()
{
int a[] = { 1, 2, 3 };
int n = sizeof(a) / sizeof(a[0]);
// Calling Function
cout << numberOfPermutations(a, n);
return 0;
} Java
// Java program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
class GFG {
// Function that returns the number of permutations
static int numberOfPermutations(int a[], int n) {
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutatio
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2 == 0) {
sumOdd += a[i];
} else {
sumEven += a[i];
}
}
// If condition holds
if (sumOdd == sumEven) {
c++;
}
} while (next_permutation(a));
// return the number of permutations
return c;
}
static boolean next_permutation(int[] p) {
for (int a = p.length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (int b = p.length - 1;; --b) {
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
// Driver Code
public static void main(String args[]) {
int a[] = {1, 2, 3};
int n = a.length;
System.out.println(numberOfPermutations(a, n));
}
}
/*This code is contributed by 29AjayKumar*/Python3
# Python3 program to find number of permutations
# such that sum of elements at odd index
# and even index are equal
def next_permutation(arr):
arrCount = len(arr);
# the head of the suffix
i = arrCount - 1;
# find longest suffix
while (i > 0 and arr[i] <= arr[i - 1]):
i-=1;
# are we at the last permutation already?
if (i <= 0):
return [False,arr];
# get the pivot
pivotIndex = i - 1;
# find rightmost element that exceeds the pivot
j = arrCount - 1;
while (arr[j] <= arr[pivotIndex]):
j-=1;
# swap the pivot with j
temp = arr[pivotIndex];
arr[pivotIndex] = arr[j];
arr[j] = temp;
# reverse the suffix
j = arrCount - 1;
while (i < j):
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i+=1;
j-=1;
return [True,arr];
# Function that returns the number
# of permutations
def numberOfPermutations(a, n):
sumEven=0;
sumOdd=0;
c = 0;
# iterate for all permutations
while (True):
# stores the sum of odd and
# even index elements
sumEven = 0;
sumOdd = 0;
# iterate for elements in permutation
for i in range(n):
# if odd index
if (i % 2):
sumOdd += a[i];
else:
sumEven += a[i];
# If condition holds
if (sumOdd == sumEven):
c+=1;
xx=next_permutation(a);
if(xx[0]==False):
break;
a=xx[1];
# return the number of permutations
return c;
# Driver Code
a = [1, 2, 3];
n = len(a);
# Calling Function
print(numberOfPermutations(a, n));
# This code is contributed by mitsC#
// C# program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
using System;
public class GFG {
// Function that returns the number of permutations
static int numberOfPermutations(int []a, int n) {
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutatio
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2 == 0) {
sumOdd += a[i];
} else {
sumEven += a[i];
}
}
// If condition holds
if (sumOdd == sumEven) {
c++;
}
} while (next_permutation(a));
// return the number of permutations
return c;
}
static bool next_permutation(int[] p) {
for (int a = p.Length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (int b = p.Length - 1;; --b) {
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
// Driver Code
public static void Main() {
int []a = {1, 2, 3};
int n = a.Length;
Console.WriteLine(numberOfPermutations(a, n));
}
}
/*This code is contributed by 29AjayKumar*/PHP
0 && $input[$i] <= $input[$i - 1])
{
$i--;
}
// are we at the last permutation already?
if ($i <= 0)
{
return false;
}
// get the pivot
$pivotIndex = $i - 1;
// find rightmost element that exceeds the pivot
$j = $inputCount - 1;
while ($input[$j] <= $input[$pivotIndex])
{
$j--;
}
// swap the pivot with j
$temp = $input[$pivotIndex];
$input[$pivotIndex] = $input[$j];
$input[$j] = $temp;
// reverse the suffix
$j = $inputCount - 1;
while ($i < $j)
{
$temp = $input[$i];
$input[$i] = $input[$j];
$input[$j] = $temp;
$i++;
$j--;
}
return true;
}
// Function that returns the number
// of permutations
function numberOfPermutations($a, $n)
{
$sumEven;
$sumOdd;
$c = 0;
// iterate for all permutations
do {
// stores the sum of odd and
// even index elements
$sumEven = $sumOdd = 0;
// iterate for elements in permutation
for ($i = 0; $i < $n; $i++)
{
// if odd index
if ($i % 2)
$sumOdd += $a[$i];
else
$sumEven += $a[$i];
}
// If condition holds
if ($sumOdd == $sumEven)
$c++;
} while (next_permutation($a));
// return the number of permutations
return $c;
}
// Driver Code
$a = array(1, 2, 3);
$n = count($a);
// Calling Function
echo numberOfPermutations($a, $n);
// This code is contributed by
// Rajput-Ji
?>输出：
2
时间复杂度： O(N！* N)