# XOR of very large Binary Numbers in range [L, R]

#include using namespace std;  string sub(string s){    int n = s.size();    for (int i = n – 1; i >= 0; i–) {                        if (s[i] == ‘0’) {            s[i] = ‘1’;        }        else {                                                s[i] = ‘0’;            break;        }    }        return s;}  string ad(string s){    int n = s.size();      int carry = 0;    for (int i = n – 1; i >= 0; i–) {                                if (s[i] == ‘1’) {            carry = 1;            s[i] = ‘0’;        }        else {                                                            carry = 0;            s[i] = ‘1’;            break;        }    }                if (carry) {        s = ‘1’ + s;    }    return s;}  string xor_finder(string s){    int n = s.size() – 1;                  int val = s[n] – ‘0’;    val = val + (s[n – 1] – ‘0’) * 2;              if (val == 0) {        return s;    }    else if (val == 1) {                        s = ‘1’;        return s;    }    else if (val == 2) {                        return (ad(s));    }    else if (val == 3) {                        s = ‘0’;        return s;    }}string final_xor(string L, string R){            while (L.size() != R.size()) {        L = ‘0’ + L;    }    string ans = “”;    for (int i = 0; i < L.size(); i++) {                                if (L[i] == '0' && R[i] == '0') {            ans += '0';        }        else if (L[i] == '0' && R[i] == '1'                 || L[i] == '1' && R[i] == '0') {                                                ans += '1';        }        else {                                                ans += '0';        }    }    return ans;}  string xorr(string L, string R){        L = sub(L);          L = xor_finder(L);          R = xor_finder(R);          string ans = final_xor(L, R);      return ans;}  int main(){        string L = "10", R = "10000";          cout