SRM 604 DIV2 500

题意：从 0，0坐标开始走，你第k（从0开始）步只能走从 x坐标或者y坐标增加 3^k的坐标，问你给定 x,y，按照这种规则能否走到

解题思路：题意可以理解为把这两个坐标的和转化为3进制，从第一位到最高位全都为1；

解题代码：

 1 // BEGIN CUT HERE
 2 /*
 3 
 4 */
 5 // END CUT HERE
 6 #line 7 "PowerOfThreeEasy.cpp"
 7 #include <cstdlib>
 8 #include <cctype>
 9 #include <cstring>
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13 #include <vector>
14 #include <string>
15 #include <iostream>
16 #include <sstream>
17 #include <map>
18 #include <set>
19 #include <queue>
20 #include <stack>
21 #include <fstream>
22 #include <numeric>
23 #include <iomanip>
24 #include <bitset>
25 #include <list>
26 #include <stdexcept>
27 #include <functional>
28 #include <utility>
29 #include <ctime>
30 using namespace std;
31 
32 #define PB push_back
33 #define MP make_pair
34 
35 #define REP(i,n) for(i=0;i<(n);++i)
36 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
37 #define FORD(i,h,l) for(i=(h);i>=(l);--i)
38 
39 typedef vector<int> VI;
40 typedef vector<string> VS;
41 typedef vector<double> VD;
42 typedef long long LL;
43 typedef pair<int,int> PII;
44 int ans[1000] = {0};
45 
46 void solve(int t)
47 {
48   
49   int k = 0 ; 
50   while(t)
51   {
52     k ++ ; 
53     ans[k] += (t % 3);
54     t = t/3;
55   }
56 }
57 class PowerOfThreeEasy
58 {
59         public:
60         string ableToGet(int x, int y)
61         {
62             memset(ans,0,sizeof(ans));
63             solve(x);
64             solve(y);
65             int i ;
66             for(i = 200;i >= 0;i --)
67                 if(ans[i] != 0 )
68                     break;
69     //        printf("%d\n\n",i);
70             for(;i >= 1; i --)
71                 if(ans[i] != 1)
72                   return "Impossible";
73           return "Possible";
74         }
75 };

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posted on 2014-01-12 14:26 dark_dream 阅读(247) 评论(0) 收藏举报