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

A. Sum of Odd Integers

题目链接-A. Sum of Odd Integers


题目大意
判断n能否表示为 k 个不同奇数之和，如果能输出YES，否则输出NO

解题思路

• 1+3+5+⋯+(2k−1)=k²，由求和公式得前k个不同奇数的和为k²，所以可得n≥k²
• 因为奇 * 奇=奇，奇 * 偶=偶，所以n，k奇偶性要相同

附上代码

#include<bits/stdc++.h>
#define int long long
#define lowbit(x) (x &(-x))
#define endl '\n'
using namespace std;
const int INF=0x3f3f3f3f;
const int dir[4][2]={-1,0,1,0,0,-1,0,1};
const double PI=acos(-1.0);
const double eps=1e-10;
const int M=1e9+7;
const int N=1e5+5;
typedef long long ll;
typedef pair<int,int> PII;
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	int t;
	cin>>t;
	while(t--){
		int n,k;
		cin>>n>>k;
		if(k>sqrt(n))
			cout<<"NO"<<endl;
		else{
			if((n-k)%2==0)
				cout<<"YES"<<endl;
			else
				cout<<"NO"<<endl;
		}
	} 
	return 0;
}