CodeForces 686C-Robbers' watch

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题意：
　　一个电子手表的示数是7进制的,现在告诉你一天有多少小时和一小时有多少分钟,问你一天里有多少个时刻,这个表上显示的数字各不相同.

分析：
　　先找出表上有多少位数字,再按位dfs,看最后得到的数是否<n和<m,把分和时转化为7进制,若位数大于7则直接输出0,若不大于零,则用dfs找到分和时的所有位不相同的情况

代码如下：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <fstream>
  5 #include <ctime>
  6 #include <cmath>
  7 #include <cstdlib>
  8 #include <algorithm>
  9 #include <set>
 10 #include <map>
 11 #include <list>
 12 #include <stack>
 13 #include <queue>
 14 #include <iterator>
 15 #include <vector>
 16 
 17 using namespace std;
 18 
 19 #define LL long long
 20 #define INF 0x3f3f3f3f
 21 #define MOD 1000000007
 22 #define MAXN 10000010
 23 #define MAXM 1000010
 24 
 25 #include <iostream>
 26 #include <cstdio>
 27 #include <cstring>
 28 #include <fstream>
 29 #include <ctime>
 30 #include <cmath>
 31 #include <cstdlib>
 32 #include <algorithm>
 33 #include <set>
 34 #include <map>
 35 #include <list>
 36 #include <stack>
 37 #include <queue>
 38 #include <iterator>
 39 #include <vector>
 40 
 41 using namespace std;
 42 
 43 #define LL long long
 44 #define INF 0x3f3f3f3f
 45 #define MOD 1000000007
 46 #define MAXN 10000010
 47 #define MAXM 1000010
 48 
 49 LL n, m;
 50 int weishu_n, zongweishu_n_m, kinds_ans;
 51 int a[10];
 52 
 53 int check()
 54 {
 55     int sum , x;
 56     sum = 0;
 57     x = 1;
 58     for(int i = weishu_n-1; i >= 0; i--)
 59     {
 60         sum += x*a[i];
 61         x *= 7;
 62     }
 63     if(sum >= n)
 64         return 0;
 65     sum = 0;
 66     x = 1;
 67     for(int i = zongweishu_n_m-1; i >= weishu_n; i--)
 68     {
 69         sum += x*a[i];
 70         x *= 7;
 71     }
 72     if(sum >= m)
 73         return 0;
 74     return 1;
 75 }
 76 
 77 void dfs(int cnt)
 78 {
 79     if(cnt == zongweishu_n_m)
 80     {
 81         if(check())
 82             kinds_ans++;
 83     }
 84     else
 85         for(int i = 0; i < 7; i++ )
 86         {
 87             int ok = 1;
 88             for(int j = 0; j < cnt; j++ )
 89                 if(i == a[j])
 90                 {
 91                     ok = 0;
 92                     break;
 93                 }
 94             if(ok)
 95             {
 96                 a[cnt] = i;
 97                 dfs(cnt+1);
 98             }
 99         }
100 }
101 
102 int main()
103 {
104     int pos;
105     int kn, km;
106     while(scanf("%lld%lld", &n, &m)==2)
107     {
108         memset(a, 0, sizeof(a));
109         kinds_ans = 0;
110         weishu_n = 0;
111         zongweishu_n_m = 0;
112         pos = 0;
113         kn = n-1;
114         if(!kn)
115             pos++;
116         while(kn)
117         {
118             kn /= 7;
119             pos++;
120         }
121         weishu_n = pos;
122         km = m-1;
123         if(!km)
124             pos++;
125         while(km)
126         {
127             km /= 7;
128             pos++;
129         }
130         zongweishu_n_m = pos;
131         dfs(0);
132         if(zongweishu_n_m > 7)
133             printf("0\n");
134         else
135             printf("%d\n", kinds_ans);
136     }
137 
138     return 0;
139 }

posted on 2016-07-16 20:23 tony-cao 阅读(168) 评论(0) 收藏举报

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