• POJ1835 宇航员 模拟


      题目链接:http://poj.org/problem?id=1835

      模拟,很无聊的题目-_-||,把转换方向映射好就可以了,我是对宇航员保存三个方向的向量,那么每次变化方向的时候就方便了。

      1 //STATUS:C++_AC_188MS_172KB
      2 #include <functional>
      3 #include <algorithm>
      4 #include <iostream>
      5 //#include <ext/rope>
      6 #include <fstream>
      7 #include <sstream>
      8 #include <iomanip>
      9 #include <numeric>
     10 #include <cstring>
     11 #include <cassert>
     12 #include <cstdio>
     13 #include <string>
     14 #include <vector>
     15 #include <bitset>
     16 #include <queue>
     17 #include <stack>
     18 #include <cmath>
     19 #include <ctime>
     20 #include <list>
     21 #include <set>
     22 #include <map>
     23 using namespace std;
     24 //define
     25 #define pii pair<int,int>
     26 #define mem(a,b) memset(a,b,sizeof(a))
     27 #define lson l,mid,rt<<1
     28 #define rson mid+1,r,rt<<1|1
     29 #define PI acos(-1.0)
     30 //typedef
     31 typedef __int64 LL;
     32 typedef unsigned __int64 ULL;
     33 //const
     34 const int N=1010;
     35 const int INF=0x3f3f3f3f;
     36 const int MOD=256,STA=8000010;
     37 const LL LNF=1LL<<60;
     38 const double EPS=1e-8;
     39 const double OO=1e15;
     40 const int dx[4]={-1,0,1,0};
     41 const int dy[4]={0,1,0,-1};
     42 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
     43 //Daily Use ...
     44 inline int sign(double x){return (x>EPS)-(x<-EPS);}
     45 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
     46 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
     47 template<class T> inline T Min(T a,T b){return a<b?a:b;}
     48 template<class T> inline T Max(T a,T b){return a>b?a:b;}
     49 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
     50 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
     51 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
     52 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
     53 //End
     54 
     55 struct Node{
     56     int x,y,z;
     57 }r1[3],r2[3];
     58 int T,n;
     59 
     60 void re(Node &a,Node b)
     61 {
     62     a.x=-b.x;
     63     a.y=-b.y;
     64     a.z=-b.z;
     65 }
     66 
     67 int main()
     68 {
     69  //   freopen("in.txt","r",stdin);
     70     int i,j,d,x,y,z;
     71     char s[20];
     72     Node *p,*q;
     73     scanf("%d",&T);
     74     while(T--)
     75     {
     76         x=y=z=0;
     77         r1[0].x=1,r1[0].y=0,r1[0].z=0;
     78         r1[1].x=0,r1[1].y=1,r1[1].z=0;
     79         r1[2].x=0,r1[2].y=0,r1[2].z=1;
     80         scanf("%d",&n);
     81         while(n--){
     82             scanf("%s%d",s,&d);
     83             if(s[0]=='f'){
     84                 r2[0]=r1[0],r2[1]=r1[1],r2[2]=r1[2];
     85             }
     86             else if(s[0]=='b'){
     87                 re(r2[0],r1[0]);
     88                 re(r2[1],r1[1]);
     89                 r2[2]=r1[2];
     90             }
     91             else if(s[0]=='l'){
     92                 re(r2[0],r1[1]);
     93                 r2[1]=r1[0];
     94                 r2[2]=r1[2];
     95             }
     96             else if(s[0]=='r'){
     97                 r2[0]=r1[1];
     98                 re(r2[1],r1[0]);
     99                 r2[2]=r1[2];
    100             }
    101             else if(s[0]=='u'){
    102                 r2[0]=r1[2];
    103                 r2[1]=r1[1];
    104                 re(r2[2],r1[0]);
    105             }
    106             else {
    107                 re(r2[0],r1[2]);
    108                 r2[1]=r1[1];
    109                 r2[2]=r1[0];
    110             }
    111 
    112             x+=r2[0].x*d;
    113             y+=r2[0].y*d;
    114             z+=r2[0].z*d;
    115             swap(r2[0],r1[0]);
    116             swap(r2[1],r1[1]);
    117             swap(r2[2],r1[2]);
    118         }
    119 
    120         int ans;
    121         if(r1[0].x==1)ans=0;
    122         else if(r1[0].y==1)ans=1;
    123         else if(r1[0].z==1)ans=2;
    124         else if(r1[0].x==-1)ans=3;
    125         else if(r1[0].y==-1)ans=4;
    126         else ans=5;
    127 
    128         printf("%d %d %d %d\n",x,y,z,ans);
    129     }
    130     return 0;
    131 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/3090473.html
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