time limit per test
1 second
memory limit per test
256 megabytes
In Sherwood, the trees are our shelter, and we are all children of the forest.
The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.
The Major Oak grows iiii new leaves in the ii-th year. It starts with 11 leaf in year 11.
Leaves last for kk years on the tree. In other words, leaves grown in year ii last between years ii and i+k−1i+k−1 inclusive.
Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year nn.
Input
The first line of the input contains a single integer tt (1≤t≤1041≤t≤104) --- the number of test cases.
Each test case consists of two integers nn, kk (1≤n≤1091≤n≤109, 1≤k≤n1≤k≤n) --- the requested year and the number of years during which the leaves remain.
Output
For each test case, output one line, "YES" if in year nn the Major Oak will have an even number of leaves and "NO" otherwise.
You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.
Example
Input
Copy
5
1 1
2 1
2 2
3 2
4 4
Output
Copy
NO
YES
NO
NO
YESNote
In the first test case, there is only 11 leaf.
In the second test case, k=1k=1, so in the 22-nd year there will be 22=422=4 leaves.
In the third test case, k=2k=2, so in the 22-nd year there will be 1+22=51+22=5 leaves.
In the fourth test case, k=2k=2, so in the 33-rd year there will be 22+33=4+27=3122+33=4+27=31 leaves.
解题说明：此题是一道数学题，每年能生成i^i个树叶，每个树叶能存活k年，求最后的数量是否是偶数。找规律能发现此题为找出以n结尾的连续整数k的和是否为偶数。
cpp
#include<iostream>
#include<cmath>
using namespace std;
void solve()
{
int n, k;
cin >> n >> k;
if (n % 2 == 0)
{
if (k / 2 % 2 == 0)
{
cout << "YES" << '\n';
return;
}
else
{
cout << "NO" << '\n';
return;
}
}
else
{
if ((k / 2 + (k % 2 != 0)) % 2 == 0)
{
cout << "YES" << '\n';
return;
}
else
{
cout << "NO" << '\n';
return;
}
}
}
int main()
{
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}