零和子序列数
给定一个由 N 个整数组成的数组 arr[] 。任务是统计总和为 0 的子序列数量。 举例:
输入: arr[] = {-1，2，-2，1} 输出: 3 所有可能的子序列为{-1，1}、{2，-2}和{-1，2，-2，1 } T7】输入: arr[] = {-2，-4，-1，6，-2} 输出: 2
方法:使用递归可以解决问题。递归地，我们从第一个索引开始，要么选择要在子序列中添加的数字，要么不选择索引处的数字。一旦索引超过 N，我们需要检查评估的和是否为 0，并且在子序列中取的数的计数应该至少为 1。如果是，那么我们简单地返回 1，加到路的数量上。 动态规划不能用来解决这个问题，因为和值可以是任何不可能存储在任何维度数组中的任何东西。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count
// of the required sub-sequences
int countSubSeq(int i, int sum, int cnt,
int a[], int n)
{
// Base case
if (i == n) {
// Check if the sum is 0
// and at least a single element
// is in the sub-sequence
if (sum == 0 && cnt > 0)
return 1;
else
return 0;
}
int ans = 0;
// Do not take the number in
// the current sub-sequence
ans += countSubSeq(i + 1, sum, cnt, a, n);
// Take the number in the
// current sub-sequence
ans += countSubSeq(i + 1, sum + a[i],
cnt + 1, a, n);
return ans;
}
// Driver code
int main()
{
int a[] = { -1, 2, -2, 1 };
int n = sizeof(a) / sizeof(a[0]);
cout << countSubSeq(0, 0, 0, a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the count
// of the required sub-sequences
static int countSubSeq(int i, int sum, int cnt,
int a[], int n)
{
// Base case
if (i == n)
{
// Check if the sum is 0
// and at least a single element
// is in the sub-sequence
if (sum == 0 && cnt > 0)
{
return 1;
}
else
{
return 0;
}
}
int ans = 0;
// Do not take the number in
// the current sub-sequence
ans += countSubSeq(i + 1, sum, cnt, a, n);
// Take the number in the
// current sub-sequence
ans += countSubSeq(i + 1, sum + a[i],
cnt + 1, a, n);
return ans;
}
// Driver code
public static void main(String[] args)
{
int a[] = {-1, 2, -2, 1};
int n = a.length;
System.out.println(countSubSeq(0, 0, 0, a, n));
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# Function to return the count
# of the required sub-sequences
def countSubSeq(i, Sum, cnt, a, n):
# Base case
if (i == n):
# Check if the Sum is 0
# and at least a single element
# is in the sub-sequence
if (Sum == 0 and cnt > 0):
return 1
else:
return 0
ans = 0
# Do not take the number in
# the current sub-sequence
ans += countSubSeq(i + 1, Sum, cnt, a, n)
# Take the number in the
# current sub-sequence
ans += countSubSeq(i + 1, Sum + a[i],
cnt + 1, a, n)
return ans
# Driver code
a = [-1, 2, -2, 1]
n = len(a)
print(countSubSeq(0, 0, 0, a, n))
# This code is contributed by mohit kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of the required sub-sequences
static int countSubSeq(int i, int sum,
int cnt, int []a, int n)
{
// Base case
if (i == n)
{
// Check if the sum is 0
// and at least a single element
// is in the sub-sequence
if (sum == 0 && cnt > 0)
{
return 1;
}
else
{
return 0;
}
}
int ans = 0;
// Do not take the number in
// the current sub-sequence
ans += countSubSeq(i + 1, sum, cnt, a, n);
// Take the number in the
// current sub-sequence
ans += countSubSeq(i + 1, sum + a[i],
cnt + 1, a, n);
return ans;
}
// Driver code
public static void Main()
{
int []a = {-1, 2, -2, 1};
int n = a.Length;
Console.Write(countSubSeq(0, 0, 0, a, n));
}
}
// This code is contributed by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the count
// of the required sub-sequences
function countSubSeq($i, $sum, $cnt, $a, $n)
{
// Base case
if ($i == $n)
{
// Check if the sum is 0
// and at least a single element
// is in the sub-sequence
if ($sum == 0 && $cnt > 0)
return 1;
else
return 0;
}
$ans = 0;
// Do not take the number in
// the current sub-sequence
$ans += countSubSeq($i + 1, $sum,
$cnt, $a, $n);
// Take the number in the
// current sub-sequence
$ans += countSubSeq($i + 1, $sum + $a[$i],
$cnt + 1, $a, $n);
return $ans;
}
// Driver code
$a = array( -1, 2, -2, 1 );
$n = count($a) ;
echo countSubSeq(0, 0, 0, $a, $n);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count
// of the required sub-sequences
function countSubSeq(i, sum, cnt, a, n)
{
// Base case
if (i == n) {
// Check if the sum is 0
// and at least a single element
// is in the sub-sequence
if (sum == 0 && cnt > 0)
return 1;
else
return 0;
}
let ans = 0;
// Do not take the number in
// the current sub-sequence
ans += countSubSeq(i + 1, sum, cnt, a, n);
// Take the number in the
// current sub-sequence
ans += countSubSeq(i + 1, sum + a[i],
cnt + 1, a, n);
return ans;
}
// Driver code
let a = [ -1, 2, -2, 1 ];
let n = a.length;
document.write(countSubSeq(0, 0, 0, a, n));
</script>
Output:
3
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