HDU 4727 The Number Off of FFF
The Number Off of FFF
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 602 Accepted Submission(s): 284
*FFF* army" is standing in a line, from left to right.
You, as the captain of
*FFF*, decides to have a "number off", that is, each soldier, from left to right, calls out a number. The first soldier should call "One", each other soldier should call the number next to the number called out by the soldier on his left side. If every soldier has done it right, they will call out the numbers from 1 to X, one by one, from left to right.
Now we have a continuous part from the original line. There are N soldiers in the part. So in another word, we have the soldiers whose id are between A and A+N-1 (1 <= A <= A+N-1 <= X). However, we don't know the exactly value of A, but we are sure the soldiers stands continuously in the original line, from left to right.
We are sure among those N soldiers, exactly one soldier has made a mistake. Your task is to find that soldier.
For each test case there are two lines. First line has the number N, and the second line has N numbers, as described above. (3 <= N <= 10
5)
It guaranteed that there is exactly one soldier who has made the mistake.
3
1 2 4
3
1001 1002 1004
Case #2: 3
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
int t,n;
long[] a;
public static void main(String[] args) throws IOException{
new Main().work();
}
void work() throws IOException{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out),true);
t=Integer.parseInt(bu.readLine()); for(int p=1;p<=t;p++){
pw.print("Case #"+p+": ");
n=Integer.parseInt(bu.readLine());
a=new long[n];
String str[]=bu.readLine().split(" ");
for(int i=0;i<n;i++){
a[i]=Long.parseLong(str[i]);
} long i=a[0];
boolean boo=true; for(int j=0;j<n;j++,i++){
if(i!=a[j]){
pw.println(j+1);
boo=false;
break;
}
}
if(boo)
pw.println(1);
}
}
}
HDU 4727 The Number Off of FFF的更多相关文章
- HDU 4727 The Number Off of FFF (水题)
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Ot ...
- HDU 4727 The Number Off of FFF 2013年四川省赛题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4727 题目大意:队列里所有人进行报数,要找出报错的那个人 思路:,只要找出序列中与钱一个人的数字差不是 ...
- 水题 HDOJ 4727 The Number Off of FFF
题目传送门 /* 水题:判断前后的差值是否为1,b[i]记录差值,若没有找到,则是第一个出错 */ #include <cstdio> #include <iostream> ...
- hdu 5898 odd-even number 数位DP
传送门:hdu 5898 odd-even number 思路:数位DP,套着数位DP的模板搞一发就可以了不过要注意前导0的处理,dp[pos][pre][status][ze] pos:当前处理的位 ...
- 周赛-The Number Off of FFF 分类: 比赛 2015-08-02 09:27 3人阅读 评论(0) 收藏
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Other ...
- hdu 2665 Kth number
划分树 /* HDU 2665 Kth number 划分树 */ #include<stdio.h> #include<iostream> #include<strin ...
- hdu 4670 Cube number on a tree(点分治)
Cube number on a tree Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/ ...
- 主席树[可持久化线段树](hdu 2665 Kth number、SP 10628 Count on a tree、ZOJ 2112 Dynamic Rankings、codeforces 813E Army Creation、codeforces960F:Pathwalks )
在今天三黑(恶意评分刷上去的那种)两紫的智推中,突然出现了P3834 [模板]可持久化线段树 1(主席树)就突然有了不详的预感2333 果然...然后我gg了!被大佬虐了! hdu 2665 Kth ...
- HDUOJ----The Number Off of FFF
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Ot ...
随机推荐
- 理解ROS话题
首先需要打开一个终端在里面运行roscore: roscore 再打开一个终端,在里面运行一个turtlesim_node节点: rosrun turtlesim turtlesim_node 打开另 ...
- websocket 通信协议
//WEBSOKET java SERVICE http://my.oschina.net/u/590484/blog/71797 UPDATE:前些天有网友mail和我讨论websocket协议,当 ...
- sql发邮件
DROP PROCEDURE USP_CheckProductCodeRepeatAndSendMail go ---检查商家是否有重复的商品编号,如果有则发送给系统配置中接收的用户邮箱 CREATE ...
- MySQL 表分区A
在MySQL中表的分区类型总的来说有四种: 第一种:range分区.基于一个给定的区间范围,把数据分配到不同的分区. 第二种:list 分区.基本枚举的值列表进行分区. 第三种:hast 分区.基 ...
- 服务器是R710常见错误汇总:
报错: E1422 CPU 1 machine check error . power cycle AC 解决方案: 系统 BIOS 已报告机器检查错误.请断开系统的交流电源 10 秒,然后重新启动系 ...
- kafka初探
http://www.infoq.com/cn/articles/kafka-analysis-part-1
- cocos2dx CCControlSlider
有的同学建议先上图,好吧,先上效果图 再看代码,创建了两个CCControlSlider在主窗口中 // on "init" you need to initialize your ...
- [poj 3159]Candies[差分约束详解][朴素的考虑法]
题意 编号为 1..N 的人, 每人有一个数; 需要满足 dj - di <= c 求1号的数与N号的数的最大差值.(略坑: 1 一定要比 N 大的...difference...不是" ...
- ViewState是什么
在做ASP.NET的时候遇到ViewState,当时不知道他是什么意思. 就在当前页面中保存数据的. 像session.是会话级别的.只要会话没有过期.session中存的数据就在. viewstat ...
- 我的IOS学习之路(三):手势识别器
在iOS的学习中,对于手势的处理是极为重要的,如对于图片,我们经常需要进行旋转,缩放以及移动等.这里做一下总结,详见代码. - (void)viewDidLoad { [super viewDidLo ...