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

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <iostream>
 #include <cmath>

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ull;
 const int INF=;
 const int maxn=+;

 ull l, r;
 string up, down;
 ull ans = ;

 int lup, ldown;

 int check(string &tmp)
 {
     if(tmp == "") return ;
     int len = tmp.size();

     if (len > lup && len < ldown) return ;
     if (len == lup && len != ldown && tmp >= up) return ;
     if (len == ldown && len != lup && tmp <= down) return ;
     if (len == lup && len == ldown && tmp >= up && tmp <= down) return ;
     return ;
 }

 int main(){

     cin >> l >> r;

     up = down = "";
     while(l)
     {
         up = (char)(l %  + '') + up;
         //cout << up << endl;
         l >>= ;
     }

     while(r)
     {
         down = (char)(r %  + '') + down;
         r >>= ;
     }

     lup = up.size(), ldown = down.size();

 //    string t;
 //    cin >> t;
 //    cout << check(t) << endl;

     for (int len = lup; len <= ldown; len++)
     {
         //cout << len << endl;
         string tmp = "";
         for (int i = ; i <= len; i++)
         {
             string one = "";
             for (int tim = ; tim <= i; tim++)
                 one += '';

             for (int j = ; j <= len; j++)
             {
                 if(i + j > len) continue;

                 string zero = "";
                 for (int tim = ; tim <= j; tim++)
                     zero += '';

                 if (len%(i+j) !=  && len%(i+j) != i) continue;

                 tmp = "";
                 int maxtim = len/(i+j);

                 for (int tim = ; tim <= maxtim; tim++)
                     tmp += one+zero;

                 if (len % (i+j) != )
                     tmp += one;

                 if (check(tmp))
                 {
                     ans++;
                     //cout << tmp << endl;
                 }

                 //cout << tmp << endl;
             }
         }

         tmp = "";
         for (int tim = ; tim <= len; tim++)
             tmp += '';
         if (check(tmp))
         {
             ans++;
             //cout << tmp << endl;
         }
     }

     cout << ans << endl;
     return ;

 }