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=2147483000;
const int maxn=200000+10;


ull l, r;
string up, down;
ull ans = 0;


int lup, ldown;

int check(string &tmp)
{
    if(tmp == "") return 0;
    int len = tmp.size();

    if (len > lup && len < ldown) return 1;
    if (len == lup && len != ldown && tmp >= up) return 1;
    if (len == ldown && len != lup && tmp <= down) return 1;
    if (len == lup && len == ldown && tmp >= up && tmp <= down) return 1;
    return 0;
}

int main(){

    cin >> l >> r;

    up = down = "";
    while(l)
    {
        up = (char)(l % 2 + '0') + up;
        //cout << up << endl;
        l >>= 1;
    }

    while(r)
    {
        down = (char)(r % 2 + '0') + down;
        r >>= 1;
    }

    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 = 1; i <= len; i++)
        {
            string one = "";
            for (int tim = 1; tim <= i; tim++)
                one += '1';

            for (int j = 1; j <= len; j++)
            {
                if(i + j > len) continue;

                string zero = "";
                for (int tim = 1; tim <= j; tim++)
                    zero += '0';

                if (len%(i+j) != 0 && len%(i+j) != i) continue;

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

                for (int tim = 1; tim <= maxtim; tim++)
                    tmp += one+zero;

                if (len % (i+j) != 0)
                    tmp += one;

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

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

        tmp = "";
        for (int tim = 1; tim <= len; tim++)
            tmp += '1';
        if (check(tmp))
        {
            ans++;
            //cout << tmp << endl;
        }
    }

    cout << ans << endl;
    return 0;

}