CodeForces 441 A. Valera and Antique Items

全栈程序员-用户IM • 2022年1月9日上午9:00 • 未分类

大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。

纯粹练JAVA….

Valera is a collector. Once he wanted to expand his collection with exactly one antique item.

Valera knows n sellers of antiques, the i-th of them auctioned ki items. Currently the auction price of the j-th object of the i-th seller issij. Valera gets on well with each of the n sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.

Unfortunately, Valera has only v units of money. Help him to determine which of the n sellers he can make a deal with.

Input

The first line contains two space-separated integers n, v (1 ≤ n ≤ 50; 104 ≤ v ≤ 106) — the number of sellers and the units of money the Valera has.

Then n lines follow. The i-th line first contains integer ki (1 ≤ ki ≤ 50) the number of items of the i-th seller. Then go ki space-separated integers si1, si2, …, siki (104 ≤ sij ≤ 106) — the current prices of the items of the i-th seller.

Output

In the first line, print integer p — the number of sellers with who Valera can make a deal.

In the second line print p space-separated integers q1, q2, …, qp (1 ≤ qi ≤ n) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.

         Sample test(s)
       

           input
         
3 50000
1 40000
2 20000 60000
3 10000 70000 190000

           output
         
3
1 2 3

           input
         
3 50000
1 50000
3 100000 120000 110000
3 120000 110000 120000

           output
         
0

Note

In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a20000 item from the second seller, and a 10000 item from the third seller.

In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him.

import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner cin=new Scanner(System.in);
		int n=cin.nextInt(),v=cin.nextInt();
		int count=0; 
		StringBuilder ans= new StringBuilder();
		for(int lll=0;lll<n;lll++)
		{
			int k=cin.nextInt();
			boolean flag=false;
			for(int i=0;i<k;i++)
			{
				int temp=cin.nextInt();
				if(temp<v)
				{
					flag=true;
				}
			}
			if(flag)
			{
				count++;
				ans.append((lll+1)+" ");
			}
		}
		System.out.println(count);
		System.out.println(ans.toString().trim());
	}
}

发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/117206.html原文链接：https://javaforall.cn

【正版授权，激活自己账号】： Jetbrains全家桶Ide使用，1年售后保障，每天仅需1毛

【官方授权正版激活】： 官方授权正版激活支持Jetbrains家族下所有IDE 使用个人JB账号...