前 N 个自然数差之和的两个可能集合为 D
原文:https://www . geesforgeks . org/可能-从第一个 n 个自然数开始的两个集合-和差为-d/
给定 N 和 D，找出是否有可能由前 N 个自然数生成两个集合，使得两个集合之和(单独)之差为 D。 示例:
Input : 5 7
Output : yes
Explanation: Keeping 1 and 3 in one set,
and 2, 4 and 5 are in other set.
Sum of set 1 = 4
Sum of set 2 = 11
So, the difference D = 7
Which is the required difference
Input : 4 5
Output : no
进场:
设 s1 和 s2 为两组。 这里我们知道 和(s1) +和(s2) = N(N+1)/2 和 和(s1)–和(s2) = D 把上面 2 个方程相加，我们得到 2 和(s1) = N(N+1)/2 + D 如果和(S1)和和(s2)都是整数，那么只有我们能把前 N 个自然数拆分成两组。为此，N(N+1)/2 + D 必须是偶数。
C++
// C++ program for implementing
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function returns true if it is
// possible to split into two
// sets otherwise returns false
bool check(int N, int D)
{
int temp = (N * (N + 1)) / 2 + D;
return (temp % 2 == 0);
}
// Driver code
int main()
{
int N = 5;
int M = 7;
if (check(N, M))
cout << "yes";
else
cout << "no";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for implementing
// above approach
class GFG
{
// Function returns true if it is
// possible to split into two
// sets otherwise returns false
static boolean check(int N, int D)
{
int temp = (N * (N + 1)) / 2 + D;
return (temp % 2 == 0);
}
// Driver code
static public void main (String args[])
{
int N = 5;
int M = 7;
if (check(N, M))
System.out.println("yes");
else
System.out.println("no");
}
}
// This code is contributed by Smitha.
Python 3
# Python program for implementing
# above approach
# Function returns true if it is
# possible to split into two
# sets otherwise returns false
def check(N, D):
temp = N * (N + 1) // 2 + D
return (bool(temp % 2 == 0))
# Driver code
N = 5
M = 7
if check(N, M):
print("yes")
else:
print("no")
# This code is contributed by Shrikant13.
C
// C# program for implementing
// above approach
using System;
class GFG
{
// Function returns true if it is
// possible to split into two
// sets otherwise returns false
static bool check(int N, int D)
{
int temp = (N * (N + 1)) / 2 + D;
return (temp % 2 == 0);
}
// Driver code
static public void Main ()
{
int N = 5;
int M = 7;
if (check(N, M))
Console.Write("yes");
else
Console.Write("no");
}
}
// This code is contributed by Ajit.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for implementing
// above approach
// Function returns true if it is
// possible to split into two
// sets otherwise returns false
function check($N, $D)
{
$temp = ($N * ($N + 1)) / 2 + $D;
return ($temp % 2 == 0);
}
// Driver code
$N = 5;
$M = 7;
if (check($N, $M))
echo("yes");
else
echo("no");
// This code is contributed by Ajit.
java 描述语言
<script>
// javascript program for implementing
// above approach
// Function returns true if it is
// possible to split into two
// sets otherwise returns false
function check( N, D)
{
let temp = (N * (N + 1)) / 2 + D;
return (temp % 2 == 0);
}
// Driver code
let N = 5;
let M = 7;
if (check(N, M))
document.write( "yes");
else
document.write("no");
// This code contributed by aashish1995
</script>
输出:
yes
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