Educational Codeforces Round 84 ( Div. 2)A. Sum of Odd Integers

You are given two integers n and k. Your task is to find if n can be represented as a sum of k distinct positive odd (not divisible by 2) integers or not.
You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤105) — the number of test cases.
The next t lines describe test cases. The only line of the test case contains two integers n and k (1≤n,k≤107).

Output
For each test case, print the answer — “YES” (without quotes) if n can be represented as a sum of k distinct positive odd (not divisible by 2) integers and “NO” otherwise.

Example
input
6
3 1
4 2
10 3
10 2
16 4
16 5
output
YES
YES
NO
YES
YES
NO

Note
In the first test case, you can represent 3 as 3.
In the second test case, the only way to represent 4 is 1+3.
In the third test case, you cannot represent 10 as the sum of three distinct positive odd integers.
In the fourth test case, you can represent 10 as 3+7, for example.
In the fifth test case, you can represent 16 as 1+3+5+7.
In the sixth test case, you cannot represent 16 as the sum of five distinct positive odd integers.

题意

有两个整数n和k，找出n是否可以表示为k个不同的正奇数的和。

思路

很显然n和k的奇偶性必须一致（奇数个奇数相加还是奇数，偶数个奇数相加是偶数）
而且k应该还有个范围，k个奇数相加的最小和（也就是前k项奇数相加的值）应该小于等于n，满足这两个条件才能输出YES
但是求前k项的和还有一个问题，如果是暴力求解的话一定会超时的，那就找找规律
1+3+5+…+(2K-1)
=(1+(2K-1))K/2
=KK
找到这个规律这个题也就解决了
注意数据类型用long long int

代码实现

#include
using namespace std;
typedef long long ll;
int main()
{
	ll t;
	ll n, k;
	cin >> t;
	while (t--)
	{
		cin >> n >> k;
		if (n < k * k) 
		{ 
			cout << "NO" << endl;
			continue;
		}
		else
		{

			if ((n % 2 == 1 && k % 2 == 1)||( n % 2 == 0 && k % 2 == 0))
				cout << "YES" << endl;
			else
				cout << "NO" << endl;
		}
	}
	return 0;
}

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