A. Marketing Scheme（思维）Educational Codeforces Round 97 (Rated for Div. 2)

Link to the original question: https://codeforces.com/contest/1437/problem/A

Test sample

input
3
3 4
1 2
120 150
output
YES
NO
YES

Note

In the first test case, you can take, for example, a=5 as the size of the pack. Then if a customer wants to buy 3 cans, he’ll buy 5 instead (3mod5=3, 52=2.5). The one who wants 4 cans will also buy 5 cans.

In the second test case, there is no way to choose a.

In the third test case, you can take, for example, a=80.

Question: Some customers want to buy xxcat food for$x$ ,$x$ belongs to the interval$[l,r]$ , and you will make a discount, which is equivalent to buying$a$ will enjoy a discount, the rest of$xmoda\; is$ purchased at the original price. Now customers have this tendency: if$x\mathrm{m}\mathrm{o}\mathrm{d}a\ge a/2$ , then customers will buy more purchases to make full discounts, of course you want customers to buy more, so please can you find a discount$a$ , so that customers always want to do this.

Question- solving ideas: Please understand the meaning of the question carefully before continuing to read the ideas. Well, let’s take a look, first discuss the condition: $xmoda\ge a/2$ , we found that if this equation holds,It is necessary to make the left big and the right small, and the way to achieve the big left is to make $a>x$ , so the remainder is always$x$ , and the small implementation method on the right is to make$a\; as$ small as possible.Then we can find $x$ max is$r$ , we can set$a$ purpose$r+1$ , then the judgment can be made. How to judge? AndxxThe value of$x$ is not determined, let’s imagine, ifxxThe minimum value of$x$ is satisfied, so is this$a$ is fine, otherwise it is not feasible. Then we can judge. One thing to note is that since it is rounding, we perform$a/2$ should be rounded up, we can also directly$(a+1)/2\; is$ fine.

AC code

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#include<bits/stdc++.h>	//POJ不支持

#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define fi first
#define se second
#define mp make_pair

using namespace std;

const int inf = 0x3f3f3f3f;//无穷大
const int maxn = 1e5;//最大值。
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>  pll;
typedef pair<int, int> pii;
//*******************************分割线，以上为自定义代码模板***************************************//

int t;
int l,r;
int main(){
    
    
	//freopen("in.txt", "r", stdin);//提交的时候要注释掉
	IOS;
	while(cin>>t){
    
    
		while(t--){
    
    
			cin>>l>>r;
			int temp=r+1;
			if(l%temp>=(temp+1)/2){
    
    
				cout<<"YES"<<endl;
			}
			else{
    
    
				cout<<"NO"<<endl;
			}
		}
	}
	return 0;
}