给定整数数组arr [] ，任务是从数组中找到大于最大元素的最小数字，而该最小数字不能使用数组中的数字来形成(将元素相加以形成其他数字)。如果不存在这样的元素，则打印-1 。
例子：
Input: arr[] = {2, 6, 9}
Output: -1
There is no such number greater than 9
that cannot be formed using 2, 6 and 9.
Input: arr[] = {6, 7, 15}
Output: 16
16 is the smallest number greater than 15 that
cannot be formed using 6, 7 and 15.
方法：该问题类似于经过最小改动的最小硬币找零问题。首先以升序对数组进行排序，然后找到最大元素max ，该最大值将是数组最后一个索引处出现的数字。检查答案的范围(最大值，2 *最大值)内的数字。如果无法使用数组的元素形成此范围内的数字，则该数字为答案，但是如果可以在此范围内形成所有数字，则不存在无法使用数组的元素形成的数字。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns the minimum
// number greater than maximum of the
// array that cannot be formed using the
// elements of the array
int findNumber(int arr[], int n)
{
// Sort the given array
sort(arr, arr + n);
// Maximum number in the array
int max = arr[n - 1];
// table[i] will store the minimum number of
// elements from the array to form i
int table[(2 * max) + 1];
table[0] = 0;
for (int i = 1; i < (2 * max) + 1; i++)
table[i] = INT_MAX;
int ans = -1;
// Calculate the minimum number of elements
// from the array required to form
// the numbers from 1 to (2 * max)
for (int i = 1; i < (2 * max) + 1; i++) {
for (int j = 0; j < n; j++) {
if (arr[j] <= i) {
int res = table[i - arr[j]];
if (res != INT_MAX && res + 1 < table[i])
table[i] = res + 1;
}
}
// If there exists a number greater than
// the maximum element of the array that can be
// formed using the numbers of array
if (i > arr[n - 1] && table[i] == INT_MAX) {
ans = i;
break;
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 6, 7, 15 };
int n = (sizeof(arr) / sizeof(int));
cout << findNumber(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function that returns the minimum
// number greater than maximum of the
// array that cannot be formed using the
// elements of the array
static int findNumber(int arr[], int n)
{
// Sort the given array
Arrays.sort(arr);
// Maximum number in the array
int max = arr[n - 1];
// table[i] will store the minimum number of
// elements from the array to form i
int table[] = new int[(2 * max) + 1];
table[0] = 0;
for (int i = 1; i < (2 * max) + 1; i++)
table[i] = Integer.MAX_VALUE;
int ans = -1;
// Calculate the minimum number of elements
// from the array required to form
// the numbers from 1 to (2 * max)
for (int i = 1; i < (2 * max) + 1; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[j] <= i)
{
int res = table[i - arr[j]];
if (res != Integer.MAX_VALUE && res + 1 < table[i])
table[i] = res + 1;
}
}
// If there exists a number greater than
// the maximum element of the array that can be
// formed using the numbers of array
if (i > arr[n - 1] && table[i] == Integer.MAX_VALUE)
{
ans = i;
break;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 6, 7, 15 };
int n = arr.length;
System.out.println(findNumber(arr, n));
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 implementation of the approach
# Function that returns the minimum
# number greater than Maximum of the
# array that cannot be formed using the
# elements of the array
def findNumber(arr, n):
# Sort the given array
arr = sorted(arr)
# Maximum number in the array
Max = arr[n - 1]
# table[i] will store the minimum number of
# elements from the array to form i
table = [10**9 for i in range((2 * Max) + 1)]
table[0] = 0
ans = -1
# Calculate the minimum number of elements
# from the array required to form
# the numbers from 1 to (2 * Max)
for i in range(1, 2 * Max + 1):
for j in range(n):
if (arr[j] <= i):
res = table[i - arr[j]]
if (res != 10**9 and res + 1 < table[i]):
table[i] = res + 1
# If there exists a number greater than
# the Maximum element of the array that can be
# formed using the numbers of array
if (i > arr[n - 1] and table[i] == 10**9):
ans = i
break
return ans
# Driver code
arr = [6, 7, 15]
n = len(arr)
print(findNumber(arr, n))
# This code is contributed by mohit kumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns the minimum
// number greater than maximum of the
// array that cannot be formed using the
// elements of the array
static int findNumber(int[] arr, int n)
{
// Sort the given array
Array.Sort(arr);
// Maximum number in the array
int max = arr[n - 1];
// table[i] will store the minimum number of
// elements from the array to form i
int[] table = new int[(2 * max) + 1];
table[0] = 0;
for (int i = 1; i < (2 * max) + 1; i++)
table[i] = int.MaxValue;
int ans = -1;
// Calculate the minimum number of elements
// from the array required to form
// the numbers from 1 to (2 * max)
for (int i = 1; i < (2 * max) + 1; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[j] <= i)
{
int res = table[i - arr[j]];
if (res != int.MaxValue && res + 1 < table[i])
table[i] = res + 1;
}
}
// If there exists a number greater than
// the maximum element of the array that can be
// formed using the numbers of array
if (i > arr[n - 1] && table[i] == int.MaxValue)
{
ans = i;
break;
}
}
return ans;
}
// Driver code
public static void Main()
{
int[] arr = { 6, 7, 15 };
int n = arr.Length;
Console.WriteLine(findNumber(arr, n));
}
}
/* This code contributed by Code_Mech */PHP
$arr[$n - 1] && $table[$i] == PHP_INT_MAX)
{
$ans = $i;
break;
}
}
return $ans;
}
// Driver code
{
$arr = array(6, 7, 15 );
$n = sizeof($arr);
echo(findNumber($arr, $n));
}
/* This code contributed by Code_Mech*/输出：
16