CodeForces 441 A. Valera and Antique Items
纯粹练JAVA....
1 second
256 megabytes
standard input
standard output
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.
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.
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.
3 50000
1 40000
2 20000 60000
3 10000 70000 190000
3
1 2 3
3 50000
1 50000
3 100000 120000 110000
3 120000 110000 120000
0
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());
}
}
版权声明:来自: 代码代码猿猿AC路 http://blog.csdn.net/ck_boss
CodeForces 441 A. Valera and Antique Items的更多相关文章
- Codeforces Round #252 (Div. 2) A - Valera and Antique Items
水题 #include <iostream> #include <set> #include <vector> #include <algorithm> ...
- Codeforces 441 B. Valera and Fruits
B. Valera and Fruits time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces441A_Valera and Antique Items(水题)
Valera and Antique Items time limit per test 1 second memory limit per test 256 megabytes input stan ...
- Codeforces#441 Div.2 四小题
Codeforces#441 Div.2 四小题 链接 A. Trip For Meal 小熊维尼喜欢吃蜂蜜.他每天要在朋友家享用N次蜂蜜 , 朋友A到B家的距离是 a ,A到C家的距离是b ,B到C ...
- Codeforces 396 E. Valera and Queries
题目链接:http://codeforces.com/problemset/problem/369/E 考虑将问题转化为有多少条线段没有覆盖这些点,如果一个询问的点集是${[x1,x2,...,xn] ...
- CodeForces - 441D: Valera and Swaps(置换群)
A permutation p of length n is a sequence of distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n). A permu ...
- Codeforces 369 C Valera and Elections
Valera and Elections 题意:现在有n个候选人, 有n-1条路, 如果选择了这个候选人, 这个候选人就会将从自己这个城市到1号城市上所有坏的路都修复一下,现在求最小的候选人数目, 如 ...
- codeforces #441 B Divisiblity of Differences【数学/hash】
B. Divisiblity of Differences time limit per test 1 second memory limit per test 512 megabytes input ...
- 【Codeforces 369C】 Valera and Elections
[链接] 我是链接,点我呀:) [题意] 给你一棵树 让你选择若干个修理点. 这些修理点被选中之后,节点i到1号节点之间的所有"坏路"都会被修好 问最少需要选择多少个点才能将所有的 ...
随机推荐
- Android kxml解析WBXML
WAP Binary XML定义好XML片断表述出同步server地址.远程数据库名称.登录账号等等内容一.两种訪问方法: 眼下的kxml支持两种wap格式:WBXML/WML. 而有两种方法将解析 ...
- 染色法判断是否是二分图 hdu2444
用染色法判断二分图是这样进行的,随便选择一个点, 1.把它染成黑色,然后将它相邻的点染成白色,然后入队列 2.出队列,与这个点相邻的点染成相反的颜色 根据二分图的特性,相同集合内的点颜色是相同的,即 ...
- Linux日志清除
因为数据要求.经常需要抓住和筛选过滤数据,大概花了7 8个月.改变了机旁数据.重新开始,发现"No space left on device" 解决方法: 直接删除日志(简单粗暴) ...
- java.lang.RuntimeException: Method called after release()
主要引起是因為在 camera.stopPreview(); camera.release(); 前沒有將setPreviewCallback 設置為null, 解決情況: public void ...
- 自己写shell命令pwd
思维:(1)得到"."的i节点号,叫n(使用stat) (2)chdir ..(使用chdir) (3)找到inode号为n的节点,得到其文件名称. 反复上述操作直到当前文件夹&q ...
- Left Menu抽屉效果ScrollView姿态共存冲突
我们有一个小项目,需要做主页ScrollView嵌套TableView. 再就是Left与Right抽屉效果. 课前有眼似的,然后完成忘. 非常糟糕的记忆.真应了那句话:好记性不如烂博客. 由于我首页 ...
- C#里System.Data.SQLite中对GUID的处理
string sqlstring = "select * from endpoint_policy where HEX([UserGuid]) ='" + CommonHelper ...
- 令人无限遐想的各种PCIe加速板卡
声明 本文不涉及不论什么特定API,也不针对不论什么特定的厂商,可是仍然值得透露一点的是,某些加速板卡厂商的成功点和失败点恰恰都是在于其通用性,在这个人们依旧依赖专业板卡的时代,依旧将板卡视为解决专业 ...
- Java Metrics
Java Metrics Java Metrics是一个功能比較强大的java统计库,它的输出组件也非常强大,帮我们做好了: 输出到Ganglia 输出到控制台 输出到JMX 输出Json 具体见:d ...
- 玩转html5(三)---智能表单(form),使排版更加方便
<!DOCTYPE html> <head> <meta http-equiv="Content-Type" content="text/h ...