求和在 [(K+1)/2, K] 范围内的子序列
给定一个由N个正整数和一个正整数K组成的数组arr[] ，任务是找到一个数组arr[]的子序列，其元素之和在[(K+1)/2, K]范围内.如果有子序列，则打印子序列的索引，否则打印-1 。
注意：这个问题可能有多个答案，打印其中任何一个。
例子：
Input: arr[ ] = {6, 2, 20, 3, 5, 6}, K = 13
Output: 0 1 3
Explanation:
Sum of elements of the subsequence {6, 2, 3} is 11 that is in the range[(K+1)/2, K].
Input: arr[ ] = {20, 24, 33, 100}, K = 2
Output: -1
方法：这个问题可以通过遍历数组arr[ ]来解决。请按照以下步骤解决此问题：
- 初始化一个向量ans来存储结果子序列的索引和totalSum来存储元素的总和 子序列。
- 使用变量i在[0, N-1]范围内迭代并执行以下步骤：
- 如果arr[i] > K ，继续遍历数组的元素。
- 如果arr[i]在[(K+1)/2, K]范围内，则返回当前元素的索引作为答案并终止循环。
- 如果arr[i] +totalSum小于K，则将当前元素添加到totalSum并在向量ans中添加索引i 。
- 如果totalSum不在给定范围内，则打印-1，否则打印 矢量答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find a subsequence of the
// given array whose sum of the elements
// is in range [K+1/2, K]
void isSumOfSubSeqInRange(int arr[], int n, int k)
{
// Vector to store the subsequence indices
vector ans;
// Variable to store the sum of subsequence
int totalSum = 0;
for (int i = 0; i < n; i++) {
// If the current element is
// greater than K then move
// forward
if (arr[i] > k) {
continue;
}
// If the current element is in
// the given range
if (arr[i] >= (k + 1) / 2) {
ans.clear();
ans.push_back(i);
totalSum = arr[i];
break;
}
// If current element and totalSum
// is less than K
else if (arr[i] + totalSum <= k) {
totalSum += arr[i];
ans.push_back(i);
}
}
// Checking if the totalSum is not
// in the given range then print -1
if (2 * totalSum < k) {
cout << -1 << endl;
return;
}
// Otherwise print the answer
for (int x : ans) {
cout << x << " ";
}
cout << endl;
}
// Driver Code
int main()
{
// Given Input
int arr[] = { 6, 2, 20, 3, 5, 6 };
int N = 6;
int K = 13;
// Function Call
isSumOfSubSeqInRange(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.Vector;
public class GFG {
// Function to find a subsequence of the
// given array whose sum of the elements
// is in range [K+1/2, K]
static void isSumOfSubSeqInRange(int arr[], int n,
int k)
{
// Vector to store the subsequence indices
Vector ans = new Vector<>();
// Variable to store the sum of subsequence
int totalSum = 0;
for (int i = 0; i < n; i++) {
// If the current element is
// greater than K then move
// forward
if (arr[i] > k) {
continue;
}
// If the current element is in
// the given range
if (arr[i] >= (k + 1) / 2) {
ans.clear();
ans.add(i);
totalSum = arr[i];
break;
}
// If current element and totalSum
// is less than K
else if (arr[i] + totalSum <= k) {
totalSum += arr[i];
ans.add(i);
}
}
// Checking if the totalSum is not
// in the given range then print -1
if (2 * totalSum < k) {
System.out.println(-1);
return;
}
// Otherwise print the answer
for (int x : ans) {
System.out.print(x + " ");
}
System.out.println();
}
// Driver code
public static void main(String[] args)
{
// Given Input
int arr[] = { 6, 2, 20, 3, 5, 6 };
int N = 6;
int K = 13;
// Function Call
isSumOfSubSeqInRange(arr, N, K);
}
}
// This code is contributed by abhinavjain194 Python3
# Python3 program for the above approach
# Function to find a subsequence of the
# given array whose sum of the elements
# is in range [K+1/2, K]
def isSumOfSubSeqInRange(arr, n, k):
# Vector to store the subsequence indices
ans = []
# Variable to store the sum of subsequence
totalSum = 0
for i in range(n):
# If the current element is
# greater than K then move
# forward
if (arr[i] > k):
continue
# If the current element is in
# the given range
if (arr[i] >= (k + 1) / 2):
ans.clear()
ans.append(i)
totalSum = arr[i]
break
# If current element and totalSum
# is less than K
elif (arr[i] + totalSum <= k):
totalSum += arr[i]
ans.append(i)
# Checking if the totalSum is not
# in the given range then print -1
if (2 * totalSum < k):
print(-1)
return
# Otherwise print the answer
for x in ans:
print(x, end = " ")
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [ 6, 2, 20, 3, 5, 6 ]
N = 6
K = 13
# Function Call
isSumOfSubSeqInRange(arr, N, K)
# This code is contributed by bgangwar59C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to find a subsequence of the
// given array whose sum of the elements
// is in range [K+1/2, K]
static void isSumOfSubSeqInRange(int[] arr, int n,
int k)
{
// Vector to store the subsequence indices
List ans = new List();
// Variable to store the sum of subsequence
int totalSum = 0;
for (int i = 0; i < n; i++) {
// If the current element is
// greater than K then move
// forward
if (arr[i] > k) {
continue;
}
// If the current element is in
// the given range
if (arr[i] >= (k + 1) / 2) {
ans.Clear();
ans.Add(i);
totalSum = arr[i];
break;
}
// If current element and totalSum
// is less than K
else if (arr[i] + totalSum <= k) {
totalSum += arr[i];
ans.Add(i);
}
}
// Checking if the totalSum is not
// in the given range then print -1
if (2 * totalSum < k) {
Console.WriteLine(-1);
return;
}
// Otherwise print the answer
foreach(int x in ans)
{
Console.Write(x + " ");
}
Console.WriteLine();
}
// Driver code
static public void Main ()
{
// Given Input
int[] arr = { 6, 2, 20, 3, 5, 6 };
int N = 6;
int K = 13;
// Function Call
isSumOfSubSeqInRange(arr, N, K);
}
}
// This Code is contributed by ShubhamSingh10 Javascript
输出
0 1 3 时间复杂度： O(N)
辅助空间： O(N)