Philosopher’s Walk --DFS
题意:
Philosopher’s Walk 图,告诉你step返回位置。

思路:
按四个块DFS
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
int abss(int a);
int lowbit(int n);
int Del_bit_1(int n);
int maxx(int a,int b);
int minn(int a,int b);
double fabss(double a);
void swapp(int &a,int &b);
clock_t __STRAT,__END;
double __TOTALTIME;
void _MS(){__STRAT=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef vector<int> VI;
typedef long long ll;
const double E=2.718281828;
const double PI=acos(-1.0);
//const ll INF=(1LL<<60);
const int inf=(<<);
const double ESP=1e-;
const int mod=(int)1e9+;
const int N=(int)1e6+; struct node
{
int x,y;
};
node dfs(int n,int step)
{
if(n==)return {,};
int block=n/;
block*=block;
if(step>=&&step<=block) return {dfs(n/,block+-step).y,n/+-dfs(n/,block+-step).x};
else if(step>block&&step<=*block) return {dfs(n/,step-block).x,n/+dfs(n/,step-block).y};
else if(step>block*&&step<=*block)return {n/+dfs(n/,step-block*).x,n/+dfs(n/,step-block*).y};
else return {n+-dfs(n/,*block+-step).y,dfs(n/,*block+-step).x};
} int main()
{
int n,step;
sc("%d%d",&n,&step);
node ans=dfs(n,step);
pr("%d %d\n",ans.x,ans.y);
return ;
} /**************************************************************************************/ int maxx(int a,int b)
{
return a>b?a:b;
} void swapp(int &a,int &b)
{
a^=b^=a^=b;
} int lowbit(int n)
{
return n&(-n);
} int Del_bit_1(int n)
{
return n&(n-);
} int abss(int a)
{
return a>?a:-a;
} double fabss(double a)
{
return a>?a:-a;
} int minn(int a,int b)
{
return a<b?a:b;
}
Philosopher’s Walk --DFS的更多相关文章
- Philosopher’s Walk(递归)
In Programming Land, there are several pathways called Philosopher’s Walks for philosophers to have ...
- Codeforces Gym 100286B Blind Walk DFS
Problem B. Blind WalkTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/cont ...
- 2017-2018 ACM-ICPC, Asia Daejeon Regional Contest Solution
A:Broadcast Stations 留坑. B:Connect3 题意:四个栈,每次放棋子只能放某个栈的栈顶,栈满不能放,现在给出(1, x) 表示黑子放在第x个栈的第一个位置,白子放在第b个栈 ...
- 2017-2018 ACM-ICPC, Asia Daejeon Regional Contest PART(10/12)
$$2017-2018\ ACM-ICPC,\ Asia\ Daejeon\ Regional\ Contest$$ \(A.Broadcast\ Stations\) \(B.Connect3\) ...
- 2017-2018 ACM-ICPC, Asia Daejeon Regional Contest F(递推)
F题 Problem F Philosopher’s Walk 题意:给你n,m,n代表一个长宽都为2的n次方的格子里,m代表走了从左下角开始走了m米,求最后的坐标. 思路: 看上图很容易便可以看出规 ...
- 2017 ACM ICPC Asia Regional - Daejeon
2017 ACM ICPC Asia Regional - Daejeon Problem A Broadcast Stations 题目描述:给出一棵树,每一个点有一个辐射距离\(p_i\)(待确定 ...
- 洛谷 P1560 [USACO5.2]蜗牛的旅行Snail Trails(不明原因的scanf错误)
P1560 [USACO5.2]蜗牛的旅行Snail Trails 题目描述 萨丽·斯内尔(Sally Snail,蜗牛)喜欢在N x N 的棋盘上闲逛(1 < n <= 120). 她总 ...
- hdu_A Walk Through the Forest ——迪杰特斯拉+dfs
A Walk Through the Forest Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/ ...
- HDU 1142 A Walk Through the Forest(最短路+dfs搜索)
A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Jav ...
随机推荐
- C语言学习笔记1-数据类型和标识符
http://blog.csdn.net/jadeshu/article/details/50751901 1.数据类型 ---1.1基本类型 --------------数值型(short(2) i ...
- 数据分析之matplotlib使用
绘制折线图 参数详情 from matplotlib import pyplot as plt # 设置图片大小,dpi图片放大缩小时可以让其更清晰 plt.figure(figsize=(20,8) ...
- HTTP header 介绍 转载
这篇文章为大家介绍了HTTP头部信息,中英文对比分析,还是比较全面的,若大家在使用过程中遇到不了解的,可以适当参考下 HTTP 头部解释 1. Accept:告诉WEB服务器自己接受什么介质类型,*/ ...
- MySQL inodb cluster部署
innodb cluster是基于组复制来实现的. 搭建一套MySQL的高可用集群innodb. 实验环境: IP 主机名 系统 软件 192.168.91.46 master RHEL7.4 mys ...
- jvm 线程状态
NEW: Just starting up, i.e., in process of being initialized.NEW_TRANS: Corresponding transition sta ...
- 字符串匹配 - hash
之前有写过一篇hash表,不过那是非常久远的时候了,应该是大一刚学一个学期的时候的成果,后来也就不那样写了,后来从xiaoxin那里学习了hash的写法,比较容易用也比较方便多hash,就这样. 分别 ...
- 7 AOP
AOP:Aspect Oriented Programming 面向切面编程.AOP是对面向对象编程的一种补充,在运行时动态地将代码切入到类的指定方法.指定位置的编程思想.将不同的方法的同一位置抽象成 ...
- js闭包小实验
js闭包小实验 一.总结 一句话总结: 闭包中引用闭包外的变量会使他们常驻内存 function foo() { var i=0; return function () { console.log(i ...
- 批量停止、删除docker容器
批量停止 根据NAMES停止所有容器 docker stop `docker ps | awk 'NR!=1{print $NF}'` 根据CONTAINER ID停止所有容器 docker stop ...
- exactly the kind of division of tasks that Gulp.js is built on
The results are then passed to a reporter function that displays the results of the code analysis in ...