Codeforces Round #767 (Div. 2)

比赛链接

Codeforces Round #767 (Div. 2)

B. GCD Arrays

time limit per test 2 seconds
memory limit per test 256 megabytes
inputstandard input
outputstandard output

Consider the array $a$ composed of all the integers in the range $[l,r]$. For example, if $l=3$ and $r=7$, then $a=[3,4,5,6,7]$.

Given $l, r$, and $k$, is it possible for $gcd(a)$ to be greater than $1$ after doing the following operation at most $k$ times?

Choose $2$ numbers from $a$.
Permanently remove one occurrence of each of them from the array.
Insert their product back into $a$.
$gcd(b)$ denotes the greatest common divisor (GCD) of the integers in $b$.

Input

The first line of the input contains a single integer $t (1≤t≤10^5)$ — the number of test cases. The description of test cases follows.

The input for each test case consists of a single line containing $3$ non-negative integers $l, r$, and $k (1≤l≤r≤10^9,0≤k≤r−l)$.

Output

For each test case, print "YES" if it is possible to have the GCD of the corresponding array greater than 1 by performing at most k operations, and "NO" otherwise (case insensitive).

Example

input

9
1 1 0
3 5 1
13 13 0
4 4 0
3 7 4
4 10 3
2 4 0
1 7 3
1 5 3

output

NO
NO
YES
YES
YES
YES
NO
NO
YES

Note

For the first test case, $a=[1]$, so the answer is "NO", since the only element in the array is $1$.

For the second test case the array is $a=[3,4,5]$ and we have $1$ operation. After the first operation the array can change to: $[3,20], [4,15]$ or $[5,12]$ all of which having their greatest common divisor equal to 1 so the answer is "NO".

For the third test case, $a=[13]$, so the answer is "YES", since the only element in the array is $13$.

For the fourth test case, $a=[4]$, so the answer is "YES", since the only element in the array is $4$.

解题思路

思维题

由于数是连续的，一些数的倍数关系，所有数的分解质因数出现的最大次数的质数是 $2$，所以只需要找出奇数出现的次数，每一次操作用偶数合并，即奇数个数减一。最终转换为奇数个数是否小于等于 $k$，另外需要特判 $l=r$ 的情况。这里求奇数个数有个公式，可转换为求偶数个数：$r/2-(l-1)/2$

• 时间复杂度：$O(t)$

代码

#include<bits/stdc++.h>
using namespace std;
int t,l,r,k;
int main()
{
    for(cin>>t;t;t--)
    {
        cin>>l>>r>>k;
        if(l==r)
            puts(l==1?"NO":"YES");
        else
            puts(r-l+1-(r/2-(l-1)/2)<=k?"YES":"NO");
    }
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15835708.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！