http://codeforces.com/problemset/problem/556/B

题意:给定n个数字且都小于n,然后每次循环第2k+1个数字+1,第2k个数字减一,k=0,1,2...n/2.  问最后能不能使n个数字刚好为排列为0,1,2,....n-1.

题解:简单模拟就好了,循环次数设为n-1,因为n次循环后回到原来的排列。

 #include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn=;
int n,a[maxn];
int main()
{
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%d",&a[i]);
int t=,sum=;
while(t<=n){
for(int i=;i<n;i++){
if(a[i]==i) sum++;
if(i%==){
if(a[i]==n-) a[i]=;
else a[i]+=;
}
else{
if(a[i]==) a[i]=n-;
else a[i]-=;
}
}
if(sum==n) break;
else sum=;
t++;
}
if(sum==n) printf("Yes\n");
else printf("No\n");
return ;
}

Codeforces Round #310 (Div. 2)--B的更多相关文章

  1. 贪心/思维题 Codeforces Round #310 (Div. 2) C. Case of Matryoshkas

    题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0 ...

  2. 构造 Codeforces Round #310 (Div. 2) B. Case of Fake Numbers

    题目传送门 /* 题意:n个数字转盘,刚开始每个转盘指向一个数字(0~n-1,逆时针排序),然后每一次转动,奇数的+1,偶数的-1,问多少次使第i个数字转盘指向i-1 构造:先求出使第1个指向0要多少 ...

  3. 找规律/贪心 Codeforces Round #310 (Div. 2) A. Case of the Zeros and Ones

    题目传送门 /* 找规律/贪心:ans = n - 01匹配的总数,水 */ #include <cstdio> #include <iostream> #include &l ...

  4. Codeforces Round #310 (Div. 1) C. Case of Chocolate set

    C. Case of Chocolate Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/ ...

  5. Codeforces Round #310 (Div. 2) B. Case of Fake Numbers 水题

    B. Case of Fake Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...

  6. Codeforces Round #310 (Div. 2) A. Case of the Zeros and Ones 水题

    A. Case of the Zeros and Ones Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/con ...

  7. Codeforces Round #310 (Div. 1) B. Case of Fugitive set

    B. Case of Fugitive Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/p ...

  8. Codeforces Round #310 (Div. 1) A. Case of Matryoshkas 水题

    C. String Manipulation 1.0 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...

  9. Codeforces Round #310 (Div. 2)--A(简单题)

    http://codeforces.com/problemset/problem/556/A 题意:给一个01字符串,把所有相邻的0和1去掉,问还剩下几个0和1. 题解:统计所有的0有多少个,1有多少 ...

随机推荐

  1. VS2005 检测内存泄漏的方法(转载)

    一.非MFC程序可以用以下方法检测内存泄露: 1.程序开始包含如下定义: #ifdef _DEBUG #define DEBUG_CLIENTBLOCK new( _CLIENT_BLOCK, __F ...

  2. Wide character in print at a2.pl line 6.

    jrhapt01:/home/tomcat/test> cat a2.pl my $str="$ARGV[0]"; use Encode; use URI::Escape; ...

  3. cssViewer牛逼的chrome插件

    很牛逼,功能很强大.

  4. 【转】调整mac电脑鼠标移动速度

    原文网址:https://www.zhihu.com/question/19579075 作者:陆猛链接:https://www.zhihu.com/question/19579075/answer/ ...

  5. SQL中and与or优先级比较

    刚刚在项目中遇到这样一个问题,SQL语句如下: select * from LOAN_BACK_LIBRARY where LIBRARY_ID=1 or LIB_ID=1 and STATUS=3 ...

  6. 动态规划(DP计数):HDU 5121 Just A Mistake

    As we all know, Matt is an outstanding contestant in ACM-ICPC. Graph problems are his favorite.Once, ...

  7. Unix环境高级编程

    1.  windows下编辑的程序怎么上传到linux下执行? //Service vsftpd status查看linux ftp服务是否开启: Linuxi ip:192.168.40.128 打 ...

  8. 页面与母版页面的asp:ContentPlaceHolder不匹配

    问题描述:The page has one or more <asp:Content> controls that do not correspond with <asp:Conte ...

  9. Poj 2299 - Ultra-QuickSort 离散化,树状数组,逆序对

    Ultra-QuickSort Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 52306   Accepted: 19194 ...

  10. GCC基本知识

    掌握下面的对GCC会有一个比较清晰的大致的了解: 不经意间,GCC已发展到了4.3的版本,尽管在软件开发社区之外乏人闻问,但因为GCC在几乎所有开源软件和自由软件中都会用到,因此它的编译性能的涨落会直 ...