Sheldon Numbers GYM -- 枚举

Sheldon Numbers GYM

题意：定义Sheldon Number为其二进制数是ABA……ABA型的或者ABAB……AB的，其中A全为1，B全为0(A>0, B>0)，问[m, n]中有多少个Sheldon Number.

思路：只需要存储好所有A与B的情况，并枚举所有的ABA……ABA型的或者ABAB……AB的情况。

# include <bits/stdc++.h>
using namespace std;

# define vi vector < int >
# define pb push_back
# define LL long long

LL n, m;
string S[3][66];
set<LL > t;

LL cal(string s){
    LL sum = 0;
    for(int i = 0; i < s.size(); i++){
        sum *= 2;
        sum += s[i]-'0';
    }
    return sum;
}

LL solve(int a, int b){
    string s;
    LL k;
    s += S[1][a];
    while(s.size()<=63){
        s += S[0][b];
        if(s.size() <=63) {
            k = cal(s);
            if(k >= n && k <= m) {
                t.insert(k);
//                cout<<k<<" "<<s<<endl;
            }
        }
        s += S[1][a];
        if(s.size() <=63) {
            k = cal(s);
            if(k >= n && k <= m) {
                t.insert(k);
//                cout<<k<<" "<<s<<endl;
            }
        }
    }
}


void init(){
    t.clear();
    for(int i = 1; i < 65; i++){
        for(int j = 0; j < i; j++){
            S[0][i] += "0";
            S[1][i] += "1";
        }
//        cout<<S[0][i]<<" "<<S[1][i]<<endl;
    }
    
    for(int i = 1; i < 63; i++){
        for(int j = 0; j < 63; j++){
            solve(i, j);
        }
    }
    cout<<t.size();
}

int main(void)
{
//    freopen("input.txt", "r", stdin);
    cin>>n>>m;
    init();
    return 0;
}