Gym - 101128H：Sheldon Numbers

题意

给你两个整数X和Y

问你在区间[X,Y]中，有多少数字的二进制满足ABAB或者A这种形式。A是某个数量的1，B是某个数量的0。

分析

因为数据规模很大，直接枚举x和y之间的数字然后判断会超时。所以直接构造符合的二进制串然后判断是不是在区间[X,Y]内就好。因为二进制串的长度最多只有63，所以随便枚举就行。但是写起来确实麻烦，容易出错的地方太多。

下面的代码是在场上两个队友写的，场上的代码嘛~可能有点乱~(好吧是我懒得再写一遍了

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <set>
  6 #include <map>
  7 #include <vector>
  8 #include <queue>
  9 #include <stack>
 10 #include <iostream>
 11 #include <cmath>
 12 
 13 using namespace std;
 14 typedef long long LL;
 15 typedef unsigned long long ull;
 16 const int INF=2147483000;
 17 const int maxn=200000+10;
 18 
 19 
 20 ull l, r;
 21 string up, down;
 22 ull ans = 0;
 23 
 24 
 25 int lup, ldown;
 26 
 27 int check(string &tmp)
 28 {
 29     if(tmp == "") return 0;
 30     int len = tmp.size();
 31 
 32     if (len > lup && len < ldown) return 1;
 33     if (len == lup && len != ldown && tmp >= up) return 1;
 34     if (len == ldown && len != lup && tmp <= down) return 1;
 35     if (len == lup && len == ldown && tmp >= up && tmp <= down) return 1;
 36     return 0;
 37 }
 38 
 39 int main(){
 40 
 41     cin >> l >> r;
 42 
 43     up = down = "";
 44     while(l)
 45     {
 46         up = (char)(l % 2 + '0') + up;
 47         //cout << up << endl;
 48         l >>= 1;
 49     }
 50 
 51     while(r)
 52     {
 53         down = (char)(r % 2 + '0') + down;
 54         r >>= 1;
 55     }
 56 
 57     lup = up.size(), ldown = down.size();
 58 
 59 //    string t;
 60 //    cin >> t;
 61 //    cout << check(t) << endl;
 62 
 63 
 64     for (int len = lup; len <= ldown; len++)
 65     {
 66         //cout << len << endl;
 67         string tmp = "";
 68         for (int i = 1; i <= len; i++)
 69         {
 70             string one = "";
 71             for (int tim = 1; tim <= i; tim++)
 72                 one += '1';
 73 
 74             for (int j = 1; j <= len; j++)
 75             {
 76                 if(i + j > len) continue;
 77 
 78                 string zero = "";
 79                 for (int tim = 1; tim <= j; tim++)
 80                     zero += '0';
 81 
 82                 if (len%(i+j) != 0 && len%(i+j) != i) continue;
 83 
 84                 tmp = "";
 85                 int maxtim = len/(i+j);
 86 
 87                 for (int tim = 1; tim <= maxtim; tim++)
 88                     tmp += one+zero;
 89 
 90                 if (len % (i+j) != 0)
 91                     tmp += one;
 92 
 93                 if (check(tmp))
 94                 {
 95                     ans++;
 96                     //cout << tmp << endl;
 97                 }
 98 
 99                 //cout << tmp << endl;
100             }
101         }
102 
103         tmp = "";
104         for (int tim = 1; tim <= len; tim++)
105             tmp += '1';
106         if (check(tmp))
107         {
108             ans++;
109             //cout << tmp << endl;
110         }
111     }
112 
113     cout << ans << endl;
114     return 0;
115 
116 }
View Code

 

posted @ 2018-04-12 00:23  蒟蒻LQL  阅读(233)  评论(0)  编辑  收藏  举报