给定数字n，将前n个自然数(1、2，… n)分成两个子集，以使两个子集的总和之间的差异最小。
例子：
Input : n = 4
Output : First subset sum = 5,
Second subset sum = 5.
Difference = 0
Explanation:
Subset 1: 1 4
Subset 2: 2 3
Input : n = 6
Output: First subset sum = 10,
Second subset sum = 11.
Difference = 1
Explanation :
Subset 1: 1 3 6
Subset 2: 2 4 5 方法：
该方法基于以下事实：通过将中间的两个元素放在一组中而将角元素放在另一组中，可以将任何四个连续的数字分为两组。因此，如果n是4的倍数，则它们的差将为0，因此一组的总和将是N个元素的总和的一半，这可以通过使用sum = n *(n + 1)/ 2来计算
在其他三种情况下，我们不能分为4组，剩下的1、2和3：
a)如果剩下的余数为1，则所有其他n-1个元素合并为4组，因此它们的总和为int(sum / 2)，另一半的总和为int(sum / 2 + 1)，并且他们的差异永远是1。
b)在n％4 == 2的情况下，也将遵循上述步骤。在这里，我们为3以后的元素形成大小为4的组。其余元素将是1和2。1进入一组，而2进入另一组。
c)当n％4 == 3时，将n-3个元素合并为4个。剩下的元素将为1、2和3，其中1和2进入一组，而3进入另一组，最终使差为0，每个集合的总和为sum / 2。
下面是上述方法的实现：
CPP
// CPP program to Minimize the absolute
// difference of sum of two subsets
#include
using namespace std;
// function to print difference
void subsetDifference(int n)
{
// summation of n elements
int s = n * (n + 1) / 2;
// if divisible by 4
if (n % 4 == 0) {
cout << "First subset sum = "
<< s / 2;
cout << "\nSecond subset sum = "
<< s / 2;
cout << "\nDifference = " << 0;
}
else {
// if remainder 1 or 2. In case of remainder
// 2, we divide elements from 3 to n in groups
// of size 4 and put 1 in one group and 2 in
// group. This also makes difference 1.
if (n % 4 == 1 || n % 4 == 2) {
cout << "First subset sum = "
<< s / 2;
cout << "\nSecond subset sum = "
<< s / 2 + 1;
cout << "\nDifference = " << 1;
}
// We put elements from 4 to n in groups of
// size 4. Remaining elements 1, 2 and 3 can
// be divided as (1, 2) and (3).
else
{
cout << "First subset sum = "
<< s / 2;
cout << "\nSecond subset sum = "
<< s / 2;
cout << "\nDifference = " << 0;
}
}
}
// driver program to test the above function
int main()
{
int n = 6;
subsetDifference(n);
return 0;
} Java
// Java program for Minimize the absolute
// difference of sum of two subsets
import java.util.*;
class GFG {
// function to print difference
static void subsetDifference(int n)
{
// summation of n elements
int s = n * (n + 1) / 2;
// if divisible by 4
if (n % 4 == 0) {
System.out.println("First subset sum = " + s / 2);
System.out.println("Second subset sum = " + s / 2);
System.out.println("Difference = " + 0);
}
else {
// if remainder 1 or 2. In case of remainder
// 2, we divide elements from 3 to n in groups
// of size 4 and put 1 in one group and 2 in
// group. This also makes difference 1.
if (n % 4 == 1 || n % 4 == 2) {
System.out.println("First subset sum = " + s / 2);
System.out.println("Second subset sum = " + ((s / 2) + 1));
System.out.println("Difference = " + 1);
}
// We put elements from 4 to n in groups of
// size 4. Remaining elements 1, 2 and 3 can
// be divided as (1, 2) and (3).
else
{
System.out.println("First subset sum = " + s / 2);
System.out.println("Second subset sum = " + s / 2);
System.out.println("Difference = " + 0);
}
}
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 6;
subsetDifference(n);
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python3 code to Minimize the absolute
# difference of sum of two subsets
# function to print difference
def subsetDifference( n ):
# summation of n elements
s = int(n * (n + 1) / 2)
# if divisible by 4
if n % 4 == 0:
print("First subset sum = ", int(s / 2))
print("Second subset sum = ",int(s / 2))
print("Difference = " , 0)
else:
# if remainder 1 or 2. In case of remainder
# 2, we divide elements from 3 to n in groups
# of size 4 and put 1 in one group and 2 in
# group. This also makes difference 1.
if n % 4 == 1 or n % 4 == 2:
print("First subset sum = ",int(s/2))
print("Second subset sum = ",int(s/2)+1)
print("Difference = ", 1)
# We put elements from 4 to n in groups of
# size 4. Remaining elements 1, 2 and 3 can
# be divided as (1, 2) and (3).
else:
print("First subset sum = ", int(s / 2))
print("Second subset sum = ",int(s / 2))
print("Difference = " , 0)
# driver code to test the above function
n = 6
subsetDifference(n)
# This code is contributed by "Sharad_Bhardwaj".C#
// C# program for Minimize the absolute
// difference of sum of two subsets
using System;
class GFG {
// function to print difference
static void subsetDifference(int n)
{
// summation of n elements
int s = n * (n + 1) / 2;
// if divisible by 4
if (n % 4 == 0) {
Console.WriteLine("First "
+ "subset sum = " + s / 2);
Console.WriteLine("Second "
+ "subset sum = " + s / 2);
Console.WriteLine("Difference"
+ " = " + 0);
}
else {
// if remainder 1 or 2. In case
// of remainder 2, we divide
// elements from 3 to n in groups
// of size 4 and put 1 in one
// group and 2 in group. This
// also makes difference 1.
if (n % 4 == 1 || n % 4 == 2) {
Console.WriteLine("First "
+ "subset sum = " + s / 2);
Console.WriteLine("Second "
+ "subset sum = " + ((s / 2)
+ 1));
Console.WriteLine("Difference"
+ " = " + 1);
}
// We put elements from 4 to n
// in groups of size 4. Remaining
// elements 1, 2 and 3 can
// be divided as (1, 2) and (3).
else
{
Console.WriteLine("First "
+ "subset sum = " + s / 2);
Console.WriteLine("Second "
+ "subset sum = " + s / 2);
Console.WriteLine("Difference"
+ " = " + 0);
}
}
}
/* Driver program to test above
function */
public static void Main()
{
int n = 6;
subsetDifference(n);
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
First subset sum = 10
Second subset sum = 11
Difference = 1