给定大小为N的整数数组arr []和整数X ，任务是计算该数组中的子序列数，以使其总和小于或等于X。
注意： 1 <= N <= 1000和1 <= X <= 1000，其中N是数组的大小。
例子：
Input : arr[] = {84, 87, 73}, X = 100
Output : 3
Explanation: The three subsequences with sum less than or equal to 100 are {84}, {87} and {73}.
Input : arr[] = {25, 13, 40}, X = 50
Output : 4
Explanation: The four subsequences with sum less than or equal to 50 are {25}, {13}, {40} and {25, 13}.
天真的方法：生成数组的所有子序列，并检查总和是否小于或等于X。
时间复杂度： O(2 N )
高效方法：使用动态编程生成子序列数。为了解决问题，请按照以下步骤操作：
- 对于任何索引IND，如果ARR [IND]≤X然后，子序列包括以及不包括当前索引处的元素的计数：
countSubsequence(ind, X) = countSubsequence(ind + 1, X) (excluding) + countSubsequence(ind + 1, X – arr[ind]) (including)
- 否则，计算不包括当前索引的子序列：
countSubsequence(ind, X) = countSubsequence(ind + 1, X) (excluding)
- 最后，从该函数返回的最终计数中减去1，因为它还会计算一个空子序列。
下面是上述方法的实现：
C++
// C++ Program to count number
// of subsequences in an array
// with sum less than or equal to X
#include
using namespace std;
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
int countSubsequenceUtil(
int ind, int sum,
int* A, int N,
vector >& dp)
{
// Base condition
if (ind == N)
return 1;
// Return if the sub-problem
// is already calculated
if (dp[ind][sum] != -1)
return dp[ind][sum];
// Check if the current element is
// less than or equal to sum
if (A[ind] <= sum) {
// Count subsequences excluding
// the current element
dp[ind][sum]
= countSubsequenceUtil(
ind + 1,
sum, A, N, dp)
+
// Count subsequences including
// the current element
countSubsequenceUtil(
ind + 1,
sum - A[ind],
A, N, dp);
}
else {
// Exclude current element
dp[ind][sum]
= countSubsequenceUtil(
ind + 1,
sum, A,
N, dp);
}
// Return the result
return dp[ind][sum];
}
// Function to return the count of subsequence
// in an array with sum less than or equal to X
int countSubsequence(int* A, int N, int X)
{
// Initialize a DP array
vector > dp(
N,
vector(X + 1, -1));
// Return the result
return countSubsequenceUtil(0, X, A,
N, dp)
- 1;
}
// Driver Code
int main()
{
int arr[] = { 25, 13, 40 }, X = 50;
int N = sizeof(arr) / sizeof(arr[0]);
cout << countSubsequence(arr, N, X);
return 0;
} Java
// Java program to count number
// of subsequences in an array
// with sum less than or equal to X
class GFG{
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
static int countSubsequenceUtil(int ind, int sum,
int []A, int N,
int [][]dp)
{
// Base condition
if (ind == N)
return 1;
// Return if the sub-problem
// is already calculated
if (dp[ind][sum] != -1)
return dp[ind][sum];
// Check if the current element is
// less than or equal to sum
if (A[ind] <= sum)
{
// Count subsequences excluding
// the current element
dp[ind][sum] = countSubsequenceUtil(
ind + 1, sum,
A, N, dp) +
// Count subsequences
// including the current
// element
countSubsequenceUtil(
ind + 1,
sum - A[ind],
A, N, dp);
}
else
{
// Exclude current element
dp[ind][sum] = countSubsequenceUtil(
ind + 1, sum,
A, N, dp);
}
// Return the result
return dp[ind][sum];
}
// Function to return the count of subsequence
// in an array with sum less than or equal to X
static int countSubsequence(int[] A, int N, int X)
{
// Initialize a DP array
int [][]dp = new int[N][X + 1];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < X + 1; j++)
{
dp[i][j] = -1;
}
}
// Return the result
return countSubsequenceUtil(0, X, A,
N, dp) - 1;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 25, 13, 40 }, X = 50;
int N = arr.length;
System.out.print(countSubsequence(arr, N, X));
}
}
// This code is contributed by Rajput-JiC#
// C# program to count number
// of subsequences in an array
// with sum less than or equal to X
using System;
class GFG{
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
static int countSubsequenceUtil(int ind, int sum,
int []A, int N,
int [,]dp)
{
// Base condition
if (ind == N)
return 1;
// Return if the sub-problem
// is already calculated
if (dp[ind, sum] != -1)
return dp[ind, sum];
// Check if the current element is
// less than or equal to sum
if (A[ind] <= sum)
{
// Count subsequences excluding
// the current element
dp[ind, sum] = countSubsequenceUtil(
ind + 1, sum,
A, N, dp) +
// Count subsequences
// including the current
// element
countSubsequenceUtil(
ind + 1,
sum - A[ind],
A, N, dp);
}
else
{
// Exclude current element
dp[ind, sum] = countSubsequenceUtil(
ind + 1, sum,
A, N, dp);
}
// Return the result
return dp[ind, sum];
}
// Function to return the count of subsequence
// in an array with sum less than or equal to X
static int countSubsequence(int[] A, int N, int X)
{
// Initialize a DP array
int [,]dp = new int[N, X + 1];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < X + 1; j++)
{
dp[i, j] = -1;
}
}
// Return the result
return countSubsequenceUtil(0, X, A,
N, dp) - 1;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 25, 13, 40 };
int X = 50;
int N = arr.Length;
Console.Write(countSubsequence(arr, N, X));
}
}
// This code is contributed by 29AjayKumar输出：
4
时间复杂度： O(N * X)